r 


REESE  LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 

,/9O       . 

No. 


succession  No.     '  £2963 


_-___,_„___- 


I ! 


I   *wg 


mis  '$ 

'M'::- 


j 

-  •      ; 


MACHINISTS' 

AND 

DRAFTSMEN'S 
HANDBOOK: 

CONTAINING  TABLES,  RULES  AND  FORMULAS, 

WITH 

NUMEROUS    EXAMPLES    EXPLAINING    THE  PRINCIPLES   OF  MATHEMATICS 
AND  MECHANICS  AS  APPLIED  TO  THE  MECHANICAL  TRADES 

INTENDED  AS  A 

REFERENCE   BOOK   FOR   ALL 

INTERESTED  IN   MECHANICAL  WORK. 


BY 

PEDER    LOBBEN, 

MECHANICAL  ENGINEER. 

Member  of  the  American  Society  of  Mechanical  Engineers  and  Worcester 

County  Mechanics  Association.     Honorary  Member  of  the  N.  A. 

Stationary    Engineers.      Non-Resident    Member    of    the 

Franklin  Institute. 


NEW  YORK: 

NOSTRAND  COMPANY, 

23  MURRAY  and  27  WARREN  STREETS. 
1900. 


COPYRIGHT,  1899, 

BY 
PEDER   LOBBEN. 


PREFACE. 


It  is  the  author's  hope  and  desire  that  this  book,  which  is 
the  outcome  of  years  of  study,  work  and  observation,  may  be 
a  help  to  the  class  of  people  to  which  he  himself  has  the  honor 
to  belong, — the  working  mechanics  of  the  world. 

This  is  not  intended  solely  as  a  reference  book,  but  it  may 
also  be  studied  advantageously  by  the  ambitious  young  engineer 
and  machinist;  and,  therefore,  as  far  as  believed  practical 
within  the  scope  of  the  work,  the  fundamental  principles  upon 
which  the  rules  and  formulas  rest  are  given  and  explained. 

The  use  of  abstruse  theories  and  complicated  formulas  is 
avoided,  as  it  is  thought  preferable  to  sacrifice  scientific  hair- 
splitting and  be  satisfied  with  rules  and  formulas  which  will 
give  intelligent  approximations  within  practical  limits,  rather 
than  to  go  into  intricate  and  complicated  formulas  which  can 
hardly  be  handled  except  by  mathematical  and  mechanical 
experts. 

In  practical  work  everyone  knows  it  is  far  more  important 
to  understand  the  correct  principles  and  requirements  of  the  job 
in  hand  than  to  be  able  to  make  elaborate  scientific  demonstra- 
tions of  the  subject;  in  short,  it  is  only  results  which  count  in 
the  commercial  world,  and  every  young  mechanic  must  remem- 
ber that  few  employers  will  pay  for  science  only.  What  they 
want  is  practical  science.  Should,  therefore,  scientific  men,  (for 
whom  the  author  has  the  greatest  respect,  as  it  is  to  the  scien- 
tific investigators  that  the  working  mechanics  are  indebted  for 
their  progress  in  utilizing  the  forces  of  nature), — find  nothing  of 
interest  in  the  book,  they  will  kindly  remember  that  the  author 
does  not  pretend  it  to  be  of  scientific  interest,  and  they  will 
therefore,  in  criticizing  both  the  book  and  the  author,  remember 
that  the  work  was  not  written  with  the  desire  to  show  the  reader 
how  vulgarly  or  how  scientifically  he  could  handle  the  subject, 
but  with  the  sole  desire  to  promote  and  assist  the  ambitious 
young  working  mechanic  in  the  world's  march  of  progress. 

P.   LOBBEN. 

NEW  YORK,  October,  1899. 


82963 


NIVERSIT 
yCS» 

IRotes  on  flfoatbematics. 


A  Unit  Js  any  quantity  represented  by  a  single  thing,  as  a 
magnitude,  or  a  number  regarded  as  one  undivided  whole. 

Numbers  are  the  measure  of  the  relation  between  quanti- 
ties of  things  of  the  same  kind  and  are  expressed  by  figures. 

Numbers  which  are  capable  of  being  divided  by  two  without 
a  remainder  are  called  even  numbers.  2?  4,  6,  8,  etc.,  are 
even  numbers. 

Numbers  which  are  not  capable  of  division  by  two  without 
giving  a  remainder  are  called  odd  numbers.  1,  3,  5,  7,  9,  etc., 
are  odd  numbers. 

A  number  which  can  not  be  divided  by  any  whole  number 
but  itself  and  the  number  1  without  giving  a  remainder  is 
called  a  prime  number.  1,  2,  3,  5,  7,  11,  13,  17,  19,  etc  ,  are 
prime  numbers. 

All  numbers  that  are  not  prime  are  said  to  be  composite 
numbers,  because  they  are  composed  of  two  or  more  factors ; 
4,  6,  8,  9,  10,  12,  etc.,  are  composite  numbers. 

Whole  numbers  are  called  integers.  Whole  numbers  are 
also  called  integral  numbers. 

A  mixed  number  is  the  sum  of  a  whole  number  and  a 
fraction. 

The  least  common  multiple  of  several  given  numbers  is 
the  smallest  number  that  can  be  divided  by  each  without  a 
remainder.  For  instance,  the  least  common  multiple  of  3,  4,  6, 
and  5  is  60,  because  60  is  the  smallest  number  that  can  be 
divided  by  those  numbers  without  a  remainder. 

Signs. 

+  (plus)  is  the  sign  of  addition. 

—  (minus  or  less),  is  the  sign  of  subtraction. 

The  signs  -f  and  —  are  also  used  to  indicate  positive  and 
negative  quantities. 

X  (times  or  multiply)  is  the  sign  of  multiplication,  but  in- 
stead of  this  sign,  sometimes  a  single  point  (.)  is  used,  especially 
in  formulas;  in  algebraic  expressions  very  frequently  factors 
are  written  without  any  signs  at  all  between  them.  For  in- 
stance, aKb  or  a.b  or  ab.  All  these  three  expressions  indicate 
that  the  quantity  a  is  to  be  multiplied  by  the  quantity  b. 


2  NOTES    ON    MATHEMATICS. 

-^-  (divided  by)  is  the  sign  of  division. 
=  (equal).     When  this  sign  is  placed  between  two  quanti- 
ties, it  indicates  that  they  are  of  equal  value.     For  instance  : 

4  +  5  +  2  =  11 


8  X  12  =  96 
100-4-    5  =  20 

.  (decimal  point)  signifies  that  the  number  written  after  it 
has  some  power  of  10  for  its  denominator. 

'      "  means  degrees,  minutes  and  seconds  of  an  angle. 

"  means  feet  and  inches. 
a'  a1'  a1"  reads  a  prime,   a  second,  a  third. 
a\  az    as  reads  a  sub  1,  a  sub  2,  a  sub  3,  and  is  always 
used  to  designate  corresponding  values  of  the  same  element. 

n  _ 

\/  This  is  the  radical  sign  and  signifies  that  a  root  is  to  be 
extracted  of  the  quantity  coming  under  the  sign  ;  this  may  be 
square  root,  cube  root,  or  any  other  root,  according  to  what 
there  is  signified  by  the  number  prefixed  in  place  of  the  letter  n. 

For  instance  :  \/reads  square  root,  \/reads  cube  root,  \/reads 

fourth  root,  V  reads  fifth  root,  \/64~=  8,  because  8  X  8  =  64 

3  _ 

V  64  =  4,  because  4  X  4  X  4  =  04 

4  _ 

V  81  =  3,  because  3X3X8X8  =  81 

The  sign  that  a  quantity  is  to  be  raised  to  a  certain 
power  is  a  small  number  placed  at  the  upper  right  hand  corner 
of  the  quantity  ;  this  number  is  called  the  exponent.  For  in- 
stance, 72  signifies  that  7  is  to  be  squared  or  multiplied  by  itself, 
that  is  : 

72  =  7  x  7  =  49 

73  =7x7X7  =  343,  etc. 

>  braces,   [   ]  brackets,  (   )  parentheses,  signify  that 

the  quantities  which  they  include  are  to  be  considered  as  one 
quantity.  For  instance  :  35  —  (8  +  6)  is  equal  to  35  —  14  =  21. 
In  this  case  the  parenthesis  indicates  that  not  only  8,  but  the 
sum  of  8  +  6  is  to  be  subtracted  from  35. 

-  (vinculum  or  bar)  is  a  straight  line  placed  over 
two  or  more  quantities,  indicating  that  they  are  to  be  operated 
upon  as  one  quantity.  For  instance,  \/25  +  11.  The  vinculum 
attached  to  the  radical  sign  indicates  that  the  square  root  shall 
be  extracted  from  the  sum  of  25+11,  which  is  the  same  as  the 
square  root  of  36. 

In  an  expression  as  —         '         the     bar     indicates     that 
3  X  b 


NOTES   ON    MATHEMATICS.  3 

the  sum  of  35+154-22  shall  be  divided  by  the  product  of  3X8 
which  is  the  same  as  72  divided  by  24. 

Whenever  a  number  or  a  quantity  is  placed  over  a  line 
and  a  number  or  a  quantity  is  placed  under  the  same  line  it 
always  indicates  that  the  number  or  quantity  over  the  line  shall 
be  divided  by  the  number  or  quantity  under  the  line.  Such  a 
quantity  is  called  a  fraction. 

The  quantity  above  the  line  is  called  the  numerator,  and 
the  quantity  below  the  line  is  called  the  denominator.  A  frac- 
tion may  be  either  proper  or  improper.  The  fraction  is  proper 
when  the  numerator  is  smaller  than  the  denominator;  for 
instance,  -?;  but  improper  if  the  numerator  is  larger  than  the 
denominator,  for  instance,  V  =  1^. 

A  fraction  can  always  be  considered  simply  as  a  problem 
in  division. 

Formulas. 

A  formula  is  an  algebraic  expression  for  some  general 
rule,  law  or  principle.  Formulas  are  used  in  mechanical 
books,  because  they  are  much  more  convenient  than  rules. 
Generally  speaking,  the  knowledge  of  algebra  is  not  required 
for  the  use  of  formulas,  because  the  numerical  values  corre- 
sponding to  the  conditions  of  the  problem  are  inserted  for 
every  letter  in  the  formula  except  the  letter  representing  the 
unknown  quantity,  which  then  is  obtained  by  simple  arithmet- 
ical calculations.  It  is  generally  most  convenient  to  begin  the 
interpretation  of  formulas  from  the  right-hand  side ;  for  instance, 
the  formula  for  the  velocity  of  water  in  long  pipes  is : 


v  =  8.02  J 


h.  d. 


f.  /. 

In  this  formula  v  represents  the  velocity  of  the  water  in 
feet  per  second. 

h  represents  the  "head"*  in  feet. 

d  represents  the  diameter  of  the  pipe  in  feet. 

f  represents  the  friction  factor  determined  by  experiments. 

/  represents  the  length  of  the  pipe  in  feet,  and  8.02  is  a 
constant  equal  to  the  square  root  ol  twice  acceleration  due  to 
gravity. 

Assume,  for  instance,  that  it  is  required  to  find  the  velocity 
of  the  flow  of  water  in  a  pipe  of  3  inches  diameter  ( %  foot ) ; 
the  length  of  the  pipe  is  1,440  feet,  the  "head"  is  9  feet,  and 
the  friction  factor  is  0.025. 

Inserting  in  the  formula  these  numerical  values,  and 
for  convenience  writing  the  diameter  of  the  pipe  in  decimals, 
we  have: 

*  In  hydraulics  the  word  "  head "  means  the  vertical  difference  between  the 
level  of  the  water  at  the  receiving  end  of  the  pipe  and  the  point  of  discharge,  or 
its  equivalent  in  pressure.  See  Hydraulics,  page  413. 


NOTES    ON    MATHEMATICS. 


v  =  8.02  XJ    9  X0.25 
* 


0.025  X  1440 
Solving  the  problem  step  by  step  we  have  : 

v  =  8.02  X  J     2-25 
>~3«' 

v  =  8.02  X  Vo.0626" 

V  —  8.02  X  0.25. 

v  =  2.005  feet  per  second. 

In  mechanical  formulas,  if  not  otherwise  specified,  it 
is  always  safe  to  assume  the  letter  g  to  mean  acceleration 
due  to  gravity,  usually  taken  as  32.2  feet  or  9.82  meters.  In 
formulas  relating  to  heat  the  letter  /  usually  signifies  the 
mechanical  equivalent  of  heat  =  778  foot  pounds  of  energy  ; 
but  in  formulas  relating  to  strength  of  materials  the  letter  J 
usually  signifies  the  polar  moment  of  inertia,  and  the  letter  / 
the  least  rectangular  moment  of  inertia.  The  letter  x  always 
expresses  the  unknown  quantity.  The  following  Greek  letters 
are  also  used  more  or  less.  The  letter  TT,  called  pi,  is  used  to 
signify  the  ratio  of  the  circumference  to  the  diameter  of  a 
circle,  and  is  usually  taken  as  3.1416.  2,  called  sigma,  usually 
signifies  the  sum  of  a  number  of  quantities.  The  letter  A, 
called  delta,  usually  signifies  small  increments  of  matter. 

The  letter  0,  called  theta,  or  the  lefter  $,  called  phi,  usually 
signifies  some  particular  angle,  sometimes  also  the  coefficient 
of  friction.  But  all  these  letters  may  be  employed  to  express 
anything,  although  it  is  usually  safe,  if  not  otherwise  specified. 
to  expect  their  meaning  to  be  as  stated.  It  is  always  customary 
to  express  known  quantities  by  the  first  letters  in  the  alphabet, 
such  as  a,  <£,  c,  etc.,  and  unknown  quantities  by  such  letters  as 
x,y,z,  etc. 


arithmetic 


Addition. 

All  quantities  to  be  added  must  be  of  the  same  unit;  we  can 
not  add  3  feet  +  8  inches  +  2  meters,  without  first  reducing 
these  three  terms  either  to  feet,  inches  or  meters.  The  same 
also  with  numbers.  Units  must  be  added  to  units,  tens  to  tens, 
hundreds  to  hundreds,  etc. 

EXAMPLE. 

318  +  5  +  38  +  10  +  H5  =  486 
Solution :        318 
5 

38 

10 

115 

486  =  Sum. 


Subtraction. 

Two  quantities  to  be  subtracted  must  be  of  the  same 
unit. 

In  subtraction,  the  same  as  in  addition,  the  units  are 
placed  under  each  other,  and  units  are  subtracted  from  units, 
tens  from  tens,  hundreds  from  hundreds,  etc. 

EXAMPLE. 

2543  —  1828  =  715 
Solution:        2543     .     .     Minuend. 

1828    .     .     Subtrahend. 

715    ..     Difference. 
Subtrahend  -}-  Difference  =  Minuend. 
(S) 


O  ARITHMETIC. 

Hultiplication. 

A  quantity  is  multiplied  by  a  number  by  adding  it  to  itself 
as  many  times'  as  the  number  indicates. 

EXAMPLE. 

314  X  3  =  314  +  314  +  314  =  942 

Solution :        314     ..     Multiplicand 
3    .    .     Multiplier 

942    .    ,     Product. 

Product 
Multiplicand       =  Multiplier. 

Product 
Multiplier        =  Multiplicand. 


Division. 

The  quantity  or  number  to  be  divided  is  called  the  dividend. 
The  number  by  which  we  divide  is  called  the  divisor.  The 
number  that  shows  how  many  times  the  divisor  is  contained  in 
the  dividend  is  called  the  quotient. 

EXAMPLE. 

6852  -4-  3  =  2284 

Solution:      3)6852(2284  6852     .     .     Dividend. 

6  3    .     .     Divisor. 

2284    .     .     Quotient. 

8  Divisor  X  Quotient  =  Dividend, 

6 

25 

24 

12 

12 


ARITHMETIC.  7 

FRACTIONS. 

Addition. 

Fractions  to  be  added  must  have  a  common  denominator  ; 
thus  we  cannot  add  %  +  %  +  X  +  3/i  unless  they  be  reduced 
to  a  common  denominator  instead  of  the  denominators  two, 
three  and  four  ;  in  other  words,  we  must  find  the  least  common 
multiple  of  the  numbers  2,  3  and  4,  which  is  12.  Thus  we 
have: 


i  =  A 

t-A 

f|  =  2A  =  2i 

EXAMPLE  2. 

Add  :      A  +  1+  J  +  A  +  I  4-  T  4-  I  +  t 

The  common  denominator  is  found  in  the  following 
manner:  Write  in  a  line  all  the  denominators,  and  divide  with 
the  prime  number,  2,  as  many  numbers  as  can  be  divided  with- 
out a  remainder.  The  numbers  that  cannot  be  divided  without 
a  remainder  remain  unchanged,  and  these  together  with  the 
quotients  of  the  divided  numbers,  are  written  in  the  next  line 
below.  Repeat  this  operation  as  long  as  more  than  one  num- 
ber can  be  divided  without  remainder,  then  try  to  divide 
by  the  next  prime  number,  and  so  on.  These  divisors  and  all 
those  numbers  remaining  undivided  in  the  last  line  are  multi- 
plied together,  and  the  product  is  the  least  common  denominator. 

2)_J0    0    *    n    9    7    p    9 
2)      0    *"?  "p    3    7    jt    9 

2)  g    ?    1       337?9 

3)  211      3?71? 


211       11713 

The  common  denominator  is  thus  : 

2X2X2X3X2X7X3=  1008 

Thus  1008  is  the  least  common  multiple  of  16,  8,  4,  12,  7 
and  9. 


8  ARITHMETIC. 

The  principle  of  this  solution  can  probably  be  better 
understood  by  resolving  these  numbers  into  prime  numbers, 
and  also  resolving  1008  into  prime  numbers  ;  we  then  find  that 
1008  contains  all  the  prime  numbers  necessary  to  make  16,  8,  4, 
12,  6,  7  and  9. 

Prime  numbers  in  1008  are  2       2       2       2       7       3       3 

"       16   "    2       2       2       2 
«  «<          «        8   "    2       2       2 

"  "          "         4   "    2       2 

"  "          "       12   "    2       2       3 

"  "          "        6   "    2       3 

"  "         "        7  "    7 

"  "         "        9  "    3       3 

Solution  of  Example  2 : 
1008 

T^  63  X  7  =  441 
|  126  X  5  =  630 

i  252  X  1  =  252 
-^2  84  X  7  =  588 
|  168  X  5  =  840 
|  144  X  5  =  720 
%  126  X  3  =  378 
£  112  X  4  =  448 


Subtraction. 

When   fractions   are  to  be  subtracted,  they  must  first  be 
reduced   to  a  common  denominator,  the  same  as  in  addition. 
EXAMPLE. 

| ;  —  J  must  be  reduced  to  |  —  i  =  I 

EXAMPLES. 

No.l.  f-^f-f^f 

No.  2.  &-i=f&-if  =  H 

No.  3.  -       =      -      = 


FRACTIONS.  9 

Multiplication. 

Fractions  are  multiplied  by  fractions,  by  multiplying  numer- 
ator by  numerator  and  denominator  by  denominator  ;  thus  : 

t  X  j\  =  f  J  =  3\ 

The  correctness  of  this  rule  can  easily  be  understood  if 
we  consider  these  two  fractions  as  two  problems  in  division. 
-*  X  i7!  will  then  be  3  divided  by  8  and  the  quotient  multiplied 
by  7  and  the  product  divided  by  12;  thus,  3  is  to  be  multiplied 
by  7  and  the  product  is  to  be  divided  by  8  times  12.  Therefore  : 

_  3  X7  1  X  7  __   r 

<  '**  ~  8  X  n  -   8X~4  -« 

A  mixed  number  may  first  be  reduced  to  an  improper  frac- 
tion and  then  multiplied  as  a  common  fraction,  numerator  by 
numerator  and  denominator  by  denominator.  For  instance: 

3|  X  f  ==  f  X  f  =  Y  ==  2f 

A  fraction  may  be  multiplied  by  a  whole  number  by  multi- 
plying the  numerator  and  letting  the  denominator  remain  un- 
changed. For  instance  : 

i7!  X  2  =  If  =  lT2f  =  U 

This  must  be  correct,  because  we  may  consider  7  as  indicat- 
ing the  quantity  and  12  as  indicating  what  kind  of  quantity  in 
exactly  the  same  sense  as  we  may  say  7  dollars  or  7  cents  ;  if 
either  of  those  were  multiplied  by  2  the  product  would,  of 
course,  be  either  dollars  or  cents  respectively,  and  for  the  same 
reason  7  twelfths  multiplied  by  2  must  be  14  twelfths. 

A  fraction  may  also  be  multiplied  by  a  whole  number,  by 
dividing  the  denominator  by  the  number  and  letting  the  numer- 
ator remain  unchanged.  For  instance  : 

Trj  X  2  =  |  '-,  —  11,  because  ^  is  equal  to  £,  so  must  T^  X  2 


EXAMPLES. 

No.  1.  31  X  f  =  V-  X  f  =  f 

No.  2.  ljxU  =  fXf  =  f 

No.  3.  AX1  =  A 

No.  4.  lXA  =  ¥X&  = 


IO  FRACTIONS. 

Division. 

A  fraction  is  divided  by  a  fraction  by  writing  the  fractions 
after  each  other,  then  inverting  the  divisor  (that  is,  changing 
its  numerator  to  denominator  and  its  denominator  to  numer- 
ator), proceed  as  in  multiplication.  For  instance : 

t  *  i'=  *  X  'fs?-ff  =  t 

The  reason  for  this  rule  can  very  easily  be  understood, 
when  we  consider  the  fractions  as  problems  in  division.  That  is 
to  say,  5  shall  be  divided  by  8  and  the  quotient  is  to  be  divided 
by  one-fourth  of  3.  But  if  the  quantity  f  is  divided  by  3  instead 
of  one-fourth  of  3,  we  must,  of  course,  multiply  the  quotient  by 
4  to  make  the  result  correct.  Therefore  : 


A  fraction  may  be  divided  by  a  whole  number  by  dividing 
the  numerator  by  the  number  and  letting  the  denominator  re- 
main unchanged.  For  instance  : 


A  fraction  may  be  divided  by  a  whole  number  by  multiply- 
ing the  denominator  by  the  whole  number  and  letting  the 
numerator  remain  unchanged.  For  instance  : 


Mixed  numbers  are  reduced  to  improper  fractions  the 
same  as  in  multiplication  ;  they  are  then  figured  the  same  as  if 
they  were  proper  fractions. 

EXAMPLES. 

No.  1. 
No.  2. 
.  No.  3. 
No.  4. 
No.  5. 

In  No.  4  it  will  be  understood  that  f  divided  by  6  must  be 
fF,  because  T^  is  exactly  a  sixth  of  $. 

In  No.  5,  also,  it  will  be  understood  that  if  -1/  is  divided  by 
4,  the  quotient  must  be  £,  because  4  is  one-fourth  of  16. 


DECIMALS.  I  I 

To  Reduce  a  Fraction  of  One  Denomination  to  a  Fraction 

of  Another  Fixed  Denomination,  and  Approx- 

imately of  the  Same  Value. 

In  mechanical  calculations,  on  drawings,  and  on  other  oc- 
casions, it  is  very  frequently  necessary  to  reduce  fractions  of 
other  denominations  to  eighths,  sixteenths,  thirty-seconds,  or 
sixty-fourths.  This  may  be  done  by  multiplying  the  numerator 
and  the  denominator  of  the  given  fraction  by  the  number  which 
is  to  be  the  denominator  in  the  new  fraction,  then  dividing  this 
new  numerator  and  denominator  by  the  denominator  of  the 
given  fraction. 

EXAMPLE. 

Reduce  %  to  eighths,  sixteenths,  thirty-seconds,  sixty- 
fourths,  or  to  hundredths. 


|f  =  -—*-  or  ft  approximately. 


f  xi 

g  =  f|  -  =  —    3L  or  }£  approximately. 

|x| 

21  '- 
|  =  AJ  =  i>7^~  or  f  1  approximately. 

I  x| 

42  2 
^i  —  }2*  —  -^2.—  or  j|-J  approximately. 

f  xi 

66  2 
•U  —  l§§  =  100-  or  xVV  approximately. 

Thus  f,  instead  of  f  ,  is  considerably  too  small,  namely, 
but  f|  is  a  great  deal  nearer,  only  T^  too  large,  and  0.67  is 
too  large. 


DECIMALS. 

In  decimal  fractions  the  denominator  is  always  some  power 
of  ten,  such  as  tenths,  hundredths,  thousandths,  etc. 

The  denominator  is  never  written,  as  it  is  fixed  by  the  rule 
that  it  is  1  with  as  many  ciphers  annexed  as  there  are  figures 
on  the  right-hand  side  of  the  decimal  point. 

yz  =  0.5  =  five-tenths  =  fV 

%  =  0.25  •=  twenty -five  hundredths  =  T^ 

y%  =  0.125  =  one  hundred  and  twenty-five  thousandths  = 

1 1^  =  1.5  =  one  and  five-tenths  =  IfV 

1#  =  1.25  =  one  and  twenty-five  hundredths  =  1T2^,  etc. 


1 2  DECIMALS. 

Figures  on  the  left  side  of  the  decimal  point  are  whole  num- 
bers. When  there  are  no  whole  numbers,  sometimes  a  cipher 
is  written  on  the  left  side  of  the  decimal  point,  but  this  is  not 
always  done,  as  it  is  common  with  many  writers  not  to  write 
anything  on  the  left  side  of  the  decimal  point  when  there  is 
no  whole  number. 
Thus: 

X  niay  be  written  .5 
X     "       "         "        .25 
%     "       "         "       .125 

It  is,  however,  preferable  to  fill  in  a  cipher  on  the  left- 
hand  side  of  the  decimal  point  when  there  is  no  whole  number, 
as  by  so  doing  the  mistake  of  reading  a  decimal  for  a  whole 
numoer  is  prevented. 

To  Reduce  a  Vulgar  Fraction  to  a  Decimal  Fraction. 

Annex  a  sufficient  number  of  ciphers  to  the  numerator, 
divide  the  numerator  by  the  denominator,  and  point  off  as  many 
decimals  in  the  quotient  as  there  are  ciphers  annexed  to  the 
numerator. 

EXAMPLE. 

Reduce  ^  to  a  decimal  fraction. 

Solution : 

8 )  7.000  ( 0.875 
64 

60 
56 

40 
40 

00 
Thus,  %  is  equal  to  the  decimal  fraction  0.875. 


DECIMALS. 


Fractions  Reduced  to  Exact  Decimals. 


A 
A 
A 

A 

.015625 
.03125 

.046875 
.0625 

tt 
A 
if 
T5* 

.265625 
.28125 
.296875 
.3125 

it 

ii 
II 

T9* 

.515625 
.53125 
.546875 
.5625 

n 
it 
« 
it 

.765625 
.78125 
.796875 
.8125 

A 

A 
A 

1 

.078125 
.09375 
.109375 
.125 

tt 

u 

II 
1 

.328125 
.34375 
.359375 
.375 

« 
i* 
II 

1 

.578125 
.59375 
.609375 
.625 

u 

H 

II 

1 

.828125 
.84375 
.859375 
.875 

A 
A 

H 
A 

.140625 
.15625 
.171875 
.1875 

3-i' 
H 
II 

TV 

.390625 
.40625 
.421875 
.4375 

tt 

§i 
If 
H 

.640625 
.65625 
.671875 
.6875 

.5  7 
<?4 

§1 
II 
H 

.890625 
.90625 
.921875 
.9375 

M 
A 

H 
i 

.203125 
.21875 
.234375 
.25 

II 
if 

tt 
i 

.453125 
.46875 
.484375 
.5 

II 
H 
tt 
1 

.703125 
.71875 
.734375 
.75 

tt 
f* 
If 
1 

.953125 
.96875 
.984375 
1. 

To  Reduce  a  Decimal  Fraction  to  a  Vulgar  Fraction. 

Write  the  decimal  as  the  numerator  of  the  fraction  and  set 
under  it  for  the  denominator  the  figure  one,  followed  by  as 
many  ciphers  as  there  are  decimal  places ;  then  cancel  the  frac- 
tion thus  written,  to  its  smallest  possible  terms. 

EXAMPLE. 

Reduce  0.3125  to  a  vulgar  fraction. 

Solution : 

0.3125  =  T^Wir*  cancelling  this  by  five  we  have  ^flfo  =  £§£ 

=  M  =  ft- 


To  Reduce  a  Decimal  Fraction  to  a  Given  Vulgar  Fraction 
of  Approximately  the  Same  Value. 

Multiply  the  decimal  by  the  number  which  is  denomi- 
nator in  the  fraction  to  which  the  decimal  shall  be  reduced, 
and  the  product  is  the  numerator  in  the  fraction. 

EXAMPLE. 

Reduce  0.4S437  to  sixteenths,  thirty-seconds  and  sixty- 
fourths. 


14  DECIMALS. 

Solution : 

0.484375  X  16  =  7.75,  gives  7^_5,  Or  T^,  approximately. 
0.484375  X  32  =  15.5,  gives  l^s,  or  ^*,  approximately. 
0.484375  X  64  =  31,  gives  |l  exactly. 

If  the  result  does  not  need  to  be  very  exact,  probably  /*, 
which  is  6\  too  small,  is  near  enough,  or  the  result,  7f  ,1%  may  be 
called  l/2,  which  is  ^  too  large.  |f  is  ^  too  small,  therefore 
either  l/2  or  |f  is  only  ^T  different  from  the  true  value.  The 
first  is  ^  too  large  and  the  last  is  -g*  too  small,  and  which 
fraction,  if  either,  should  be  preferred,  will  depend  entirely  upon 
the  purpose  for  which  the  problem  is  solved.  ^  is  the  exact 
value. 

Addition  of  Decimal  Fractions. 

In  adding  decimal  fractions,  care  should  be  taken  to  place 
the  decimal  points  under  each  other  ;  then  add  as  if  they  were 
whole  numbers. 

EXAMPLE. 

Add  50.5  +  5.05  +  0.505  +  0.0505 
Solution  : 

50.5 
5.05 
0.505 
0.0505 

56.1055 

To  prevent  mistakes  and  mixing  up  of  the  figures  during 
addition,  it  is  preferable  to  make  all  the  decimal  fractions  in 
the  problem  of  the  same  denomination  by  annexing  ciphers. 
Thus :  50.5000 

5.0500 
0.5050 
0.0505 
56.1055 

Subtraction  of  Decimal  Fractions. 

The  decimal  point  in  the  subtrahend  must  be  placed  under 
that  in  the  minuend  ;  the  fractions  are  both  brought  to  the 
same  denomination  by  annexing  ciphers,  then  the  subtraction 
is  performed  just  as-  if  they  were  whole  numbers,  but  close 
attention  must  be  paid  to  have  the  decimal  point  in  the  same 
place  in  the  difference  as  it  is  in  the  minuend  and  subtrahend. 

EXAMPLE. 

318.05  —  121.6542 


DECIMALS.  1 5 

Solution :  318.0500     Minuend. 

121.6542     Subtrahend. 
196.3958     Difference. 

riultiplication  of  Decimal  Fractions. 

Multiply  the  factors  as  if  they  were  whole  numbers.  After 
multiplication  is  performed,  count  the  number  of  decimals  in 
both  multiplier  and  multiplicand  and  point  off  (from  the  right) 
the  same  number  of  decimals  in  the  product. 

If  there  are  not  enough  figures  in  the  product  to  give  as 
many  decimals  as  required,  then  prefix  ciphers  on  the  left  until 
the  required  number  of  decimals  is  obtained. 

EXAMPLE  1. 

0.08  X  O.OG5  =  0.00520  =  0.0052 

In  this  example  it  is  necessary  after  the  multiplication  is 
performed,  to  prefix  two  ciphers  to  the  product  in  order 
to  obtain  the  necessary  number  of  decimals,  because  the  pro- 
duct, 520,  consists  of  only  three  figures,  but  the  two  numbers, 
0.08  and  0.065,  contain  five  decimals. 

EXAMPLE  No.  2. 

3.1416  X  5  =  15.7080  =  15.708 

EXAMPLE  No.  3. 

3.1416  X  0.5  =  1.57080  =  1.5708 

Division  of  Decimal  Fractions. 

Divide  same  as  in  whole  numbers,  and  point  off  in  the 
quotient  as  many  decimals  as  the  number  of  decimals  in  the 
dividend  exceeds  the  number  of  decimals  in  the  divisor. 

If  the  divisor  contains  more  decimals  than  the  dividend, 
then  before  dividing  annex  ciphers  (on  the  right-hand  side)  in 
the  dividend  until  dividend  and  divisor  are  both  of  the  same 
denomination,  then  the  quotient  will  be  a  whole  number. 

EXAMPLE. 

43.62  -r-  0.003  —  14,540 

Solution  :  0.003  )  43.620  (  14,540 

3 

13 
12 

16 
15 

12 
12 

00 


1 6  RATIO    AND    PROPORTION. 

In  this  example  the  dividend  consists  of  only  two  decimals, 
but  the  divisor  has  three,  therefore  we  have  to  annex  a  cipher 
to  the  dividend.  This  brings  divisor  and  dividend  to  the  same 
denomination,  and  the  quotient  is  a  whole  number. 

EXAMPLE  2. 

43.62  ^-  0.3  =  145.4 

In  this  example  the  dividend  has  one  decimal  more  than 
the  divisor,  therefore  the  quotient  has  one  decimal. 


RATIO. 

The  word  ratio  causes  considerable  ambiguity  in  mechani- 
cal books,  as  it  is  frequently  used  with  different  meaning  by 
different  writers. 

The  common  understanding  seems  to  be  that  the  ratio  be- 
tween two  quantities  is  the  quotient  when  the  first  quantity  is 
divided  by  the  last  quantity ;  for  instance,  the  ratio  between  3 
and  12  is  X>  but  the  ratio  between  12  and  3  is  4.  The  ratio  be- 
tween the  circumference  of  a  circle  and  its  diameter  is  TT  or 
8.1410,  but  the  ratio  between  the  diameter  and  the  circumfer- 
ence is  ^-  or  0.3183,  etc.  This  is  the  sense  in  which  the  word  is 
used  in  this  book,  as  this  seems  to  agree  with  the  common  cus- 
tom with  most  mechanical  writers. 

The  term  ratio  is  also  sometimes  applied  to  the  difference 
of  two  quantities  as  well  as  to  their  quotient ;  in  which  case  the 
former  is  called  arithmetical  ratio,  and  the  latter  geometrical 
ratio.  (See  Progressions,  page  68.) 


PROPORTION. 

In  simple  proportion  there  are  three  'known  quantities  by 
which  wre  are  able  to  find  the  fourth  unknown  quantity  ;  there- 
fore proportion  is  also  called  "the  rule  of  three",  and  it  is  either 
direct  or  inverse  proportion. 

It  is  called  direct  proportion  if  the  terms  are  in  such  ratio  to 
one  another  that  if  one  is  doubled  then  the  other  will  also  have 
to  be  doubled,  or  if  one  is  halved  the  other  must  also  be  halved. 
For  instance,  if  50  pounds  of  steel  cost  $25,  how  much  will  250 
pounds  cost? 

50  Ibs.  cost  $25;  250  must  cost  -50  X  — '  =  $125. 

50 

This  is  direct  proportion,  because  the  more  steel  we  buy, 
the  more  money  we  have  to  pay. 

In  inverse  proportion  the  terms  are  in  such  ratio  that  if  one 
is  doubled  the  other  is  halved,  or  if  one  is  halved  the  other  is 
doubled. 


PROPORTION.  I  7 

EXAMPLE. 

Eight  men  can  finish  a  certain  work  in  12  days.  How  many 
men  are  required  to  do  the  same  work  in  3  days  ? 

Here  we  see  that  the  fewer  days  in  which  the  work  is  to  be 
done,  the  more  men  are  required.  Therefore,  this  example  is 
in  inverse  proportion. 

In  12  days  the  work  was  done  by  8  men ;  therefore,  in  order 

to  do  the  work  in  3  days  it  will  require  -      L2  =  32  men. 

It  requires  4  times  as  many  men  because  the  work  is  to  be 
done  in  one  quarter  of  the  time. 

Compound  Proportion. 

A  proportion  is  called  compound,  if  to  the  three  terms  there 
are  combined  other  terms  which  must  be  taken  into  considera- 
tion in  solving  the  problem. 

A  very  easy  way  to  solve  a  compound  proportion  is  to 
(same  as  is  shown  in  the  following  examples)  place  the  .con- 
ditional proposition  under  the  interrogative  sentence,  term 
for  term,  and  write  x  for  the  unknown  quantity  in  the  inter- 
rogative sentence ;  draw  a  vertical  line ;  place  x  at  the  top  at  the 
left-hand  side ;  then  try  term  for  term  and  see  if  they  are  direct 
or  inverse  proportionally  relative  to  .r,  exactly  the  same  way  as 
if  each  term  in  the  conditional  proposition  and  the  correspond- 
ing term  in  the  interrogative  sentence  were  terms  in  a  simple 
rule-of-three  problem.  Arrange  each  term  in  the  interrogative 
sentence  either  on  the  right  or  left  of  the  vertical  line,  according 
to  whether  it  is  found  to  be  either  a  multiplier  or  a  divisor,  when 
the  problem,  independent  of  the  other  terms,  is  considered  as  a 
simple  rule-of-three  problem. 

After  all  the  terms  in  the  interrogative  sentence  are  thus 
arranged,  place  each  corresponding  term  in  the  conditional 
proposition  on  the  opposite  side  of  the  vertical  line.  Then 
clear  away  all  fractions  by  reducing  them  to  improper  frac- 
tions, and  let  the  numerator  remain  on  the  same  side  of  the  verti- 
cal line  where  it  is,  but  transfer  the  denominator  to  the  opposite 
side.  Now  cancel  any  term  with  another  on  the  opposite 
side  of  the  vertical  line;  then  multiply  all  the  quantities  on  the 
right  side  of  the  vertical  line  with  each  other.  Also  multiply 
all  the  quantities  on  the  left  side  of  the  vertical  line  with  each 
other. 

Divide  the  product  on  the  right  side  by  the  product  on  the 
left,  and  the  quotient  is  the  answer  to  the  problem, 

EXAMPLE  1. 

A  certain  work  is  executed  by  15  men  in  G  days,  by  work- 
ing 8  hours  each  day.  How  many  days  would  it  take  to  do  the 
same  amount  of  work  if  12  men  are  working  7^  hours  each 
day? 


PROPORTION. 


Solution : 


15  Men  6  Days  8     Hours. 

12    "    x    "     i2  " 


ty* 


8  Days. 


EXAMPLE  2. 


A  steam  engine  of  25  horse  power  is  using  1500  pounds  of 
coal  in  1  day  of  9^  working  hours.  How  many  pounds  of  coal 
in  the  same  proportion  will  be  required  for  2  steam  engines  each 
having  30  horse  power,  working  6  days  of  12%  hours  each  day  ? 


Solution : 

1  Machine  25  Hp.  1,500  pounds  1  Day 

2  "         30  x         "6      " 

x     \  im    300 
1  2 


hours. 


28,800  pounds  of  coal. 


EXAMPLE  3. 


A  piece  of  composition  metal  which  is  12  inches  long,  %l/2 
inches  thick  and  4>£  inches  wide,  weighs  45  pounds.  How  many 
pounds  will  another  piece  of  the  same  alloy  weigh,  if  it  meas- 
ures 8  inches  long,  1%  inches  thick  and  6%  inches  wide? 


12" 

8 

2 
1 

long,  3X" 
.r 

7         %y2 

$         £/4 
l         * 

1         £ 

thick,  4^"  wide, 
"      6%        " 
45 
?           ? 
I3X      / 

?'4       1 
?         1 

45  pounds. 

x        " 

1 

2 

45 

22^  pounds. 

INTEREST.  19 

INTEREST. 

The  money  paid  for  the  use  of  borrowed  capital  is  called 
interest.  It  is  usually  figured  by  the  year  per  100  of  the 
principal. 

Simple  Interest. 

Simple  interest  is  computed  by  multiplying  the  principal  by 
the  percentage,  by  the  time,  and  dividing  by  100. 

What  is  the  interest  of  $125,  for  3  years,  at  4%  per  year? 
Solution  : 

125  X  4  X  3  =  £15 
100 

In  Table  No.  1,  under  the  given  rate  per  cent.,  find  the 
interest  for  the  number  of  years,  months,  and  days-;  add  these 
together,  and  multiply  by  the  principal  invested,  and  the 
product  is  the  interest. 

EXAMPLE. 

What  is  the  interest  of  $600,  invested  at  6%,  in  5  years, 
3  months,  and  (>  days? 

Solution : 

$1.00  in  5  years      at  6%  =  0.30 
"      "  3  months  "    "     =  0.015 
"      "  6  days        "    "     =  0.001 


0.316 
600  =  Principal. 


$189.60  =  Interest 


20 


INTEREST. 


>> 

I 

«} 

1 

o 

I 


Ci  C5  CO  GO  IT—  t—  O?O»C^'^ICCCC 


oooooooooooooooooooooo 
o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o 


O  *"^  C^  CC  "^  *O  ^O  1^*  CO  O  T~H  C^  CC  "^  *O  CO  J^*  QC  O  ^"  (T^J 

o  o  o  0  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o  o 


"•  «  PO  ^  IC^O  tNOO  O\  O  •• 


INOO  a  o  - 


t>.oo 


INTEREST. 


21 


o    o^ 


ooooooooooo 

._  I  if?  O  ».t  O  »(t  O  O  O  »O  O  »f? 

* Islliiillsi 


- 

20     O  <-H  <M  (7<l  Ct  ^  -t  »^   C: 
00     COOOOOOO'O 


77 


CO  t—  ©  CO  l^"  ©  CO  1—  ©  CO  t~* 


-t  »t>  »O 

000 


05  *••  5>  co  *-  Q  oo 

5f  -2  S  TO  S  S  co 


OT^Olr-O'NOt^ 

T_lT_t^-(^-((M/NiMfN 
OOOOOOOO 


^  IftO  l>00  O\  O  - 


^ 


oooooooooo 

i-i'NCO-^iOsOl-OOOlO 


O  O  O  »O  O  >O  O  »O  O 
r-l'FH<N<MCOCO'<*^t|ia 


COCOCStMlOGC^H^t-O 
OOOi-lr-lT-t<N<N<NCC 


ISOO  Os  O 


22  COMPOUND    INTERESt. 

Compound  Interest  Computed  Annually. 

If  the  interest  is  not  withdrawn,  but  added  to  the  principal, 
so  that  it  will  also  draw  interest,  it  is  called  compound  interest. 

EXAMPLE. 

What  is  the   amount  of  #300,  in  3  years,   at  ~>%  ?     The 
interest  is  added  to  the  principal  at  the  end  of  each  year. 

Solution : 

Principal  and  interest  at  the  end  of  first  year, 

105  XJ 
100 

Principal  and  interest  at  the  end  of  second  year, 
105  X  315  -  #330.75. 

Principal  and  interest  at  the  end  of  third  year, 
105  X  330.75 


100 


=  #347.2875,  =  #347.29  =  Amount. 


When  compound  interest  for  a  great  number  of  years  is  to  be 
calculated,  the  above  method  of  figuring  will  take  too  mucli 
time,  and  the  following  interest  tables,  No.  2  and  No.  3,  are 
computed  in  order  to  facilitate  such  calculations. 

In  Table  No.  2,  under  the  given  rate  per  cent.,  and  opposite 
the  given  number  of  years,  find  the  amount  of  one  dollar  in- 
vested at  that  rate  for  the  time  taken.  Multiply  this  by  the 
principal  invested  and  the  product  is  the  amount. 

EXAMPLE. 

#400  is  invested  at  5%  compound  interest  for  17  years,  com- 
puted annually.  What  is  the  amount? 

Solution : 

In  Table  No.  2,  under  5%,  and  opposite  17  years,  we  find 
2.292011.  Multiply  this  by  the  principal. 

Thus: 

2.292011 
400 


916.8044  =  #910.80  =  Amount. 


COMPOUND    INTEREST. 


COCO'Ni-l<Mt-OCOCOCOr-CO~tiC5i-i»<OT-iXr-t—  »OCOO 
-tCJiOCOXXCOOiOCi-^COCOOi-iCOCOXCOOCOI—  "* 
OI~-»Or-i"-i-iTfi-lXT-iX»OOT-lXO5OCOiO- 


^  O*  CO  CO  ^^ 


»o 


o  i-  «  c  10 


Ot  iO  Ol  O  Ci  O  T—H  ~^  Cl 

S  55  o  S  c  «  a  o  55 


CB 

1=1 


I 


^ 


r- 

t—  »O  sO  O  t—  I"-  *->  CO  O  >.1  O  Ci  'X  'M 

»Oi-Ht^TfOt~>O'Mi—  iCt'Xl'-l'-X 


«  04  04  <N  <N  09  04  Qf9  00  00  00 


§>, 

tt 

~J 

^E 

O  3 

*-  c 

g* 
Is 

p 

£>> 


IS 

J= 
^ 


^ 


"^  ^  1"^*  Oi  ^H  ^  i^  CO  ^^  i™^ 


COCOCi( 
l--OClCiCOCOOC5OGOr-  OOO500CNOO 

t—  O  CO  -M  7-1  X  X  ».^  Cl  O  Ci  CO  CO  Cl  CO  -f  Tft  J^  Tf  O  ^H 

O"*!—  (NOSOf-HCO 


l^C:-tOCOCiCO>CCOO'MXCOOCOT-i-tOfN—  <-t—  'O 
ri  C:  I-  O  1-  CO  i-  —  i  CO  '-C  CO  X  CO  C:  -f  CO  O  O  C:  O  00  d  1- 
»ioWgoOt«»r»G6G8t»iQK50&t»00^»iH  01  —  »r?  t^  t^ 

o  co  o  co  co  01  co 
O  co  —  t-  co  c: 
-  aq  GO  cs  cs  q  q 


o  01  »c  c:  -t  c:  to  -t  co  -t  >.o  x  n  i  —  f  T-I 

co  co  c:  -M  o  c:  01  co  O  -f  x  01  -^  —  o  o  o 
c  -  —  —  01  01  co  co  co  -t  -t  -o  o  to  to 


\N  »o  o  to  co  —  a  x  x 
•»\  <M  ic  r-  o  co  o  x  — 


o  - 


o  - 


24  COMPOUND    INTEREST. 

Compound  Interest  Computed  Semi=AnnualIy. 

When  compound  interest  is  to  be  computed  semi-annually,. 
use  Table  No.  3.  Under  the  given  rate  and  opposite  the  given 
number  of  years,  find  the  amount  of  one  dollar  invested  and 
interest  computed  semi-annually  for  the  time  taken.  Multiply 
this  by  the  principal  invested,  and  the  product  is  the  amount. 

EXAMPLE. 

$350  is  put  in  a  savings  bank  paying  4%,  computed  semi- 
annually.  What  is  the  amount  in  10  years  ? 

Solution : 

Under  4%,  and  opposite  10  years,  we  find  the  number 
1.4860.  This  we  multiply  by  the  principal  invested. 

Thus: 

1.485949 
350 


520.08215  =  $520.08  =  Amount. 

To  compute  compound  interest  for  longer  time  than 
is  given  in  the  tables,  figure  the  amount  for  as  long  a  time 
as  the  table  gives ;  then  consider  this  amount  as  a  new  princi- 
pal invested,  and  use  the  table  and  figure  again  for  the  rest  of 
the  time. 

EXAMPLE. 

What  is  the  amount  of  $40,  left  in  a  savings  bank  18  years, 
at  4%,  and  the  interest  computed  semi-annually.  The  table 
only  gives  12  years,  therefore  we  will  look  opposite  12  years, 
under  4%,  and  find  the  number  1.608440.  This  we  multiply  by 
the  principal  invested. 

Thus: 

1.608440 
40 


64.3376 

But  now  we  have  to  compute  for  6  years  more,  therefore 
under  4%,  and  opposite  6  years,  we  find  the  number  1.2(58243. 
Multiplying  this  by  the  principal,  which  is  now  considered  as 
being  invested  6  years  more,  we  have : 

1.268243  X  64.3376  =  $81.60  =  Amount. 

Thus,  $40,  invested  at  4%  interest,  computed  semi-annually, 
will,  after  18  years  of  time,  amount  to  $81.60. 


COMPOUND    INTEREST. 


o  %  3  22 


^ 

1/5 


(M^COX 

O  O  O  O 


>sQ 


G^  1-  ?C  X  -t  -t  C: 

N»co*M^TCpTt<»j- 

o  o  >o  T-I  i—  co  cr.  --c  co  o  t—  »ff  co  *-i 

T-iCO^t^5i~-CtO5<lTfOt-OS^CO 
O  O  O         rH  1-1  I-H  »^  I-H  -H  (>1  7-1 


»O(?:lXCO'*f 
X  Ci  74  'N  >C 


\N 


1-1  ~t 
t^*  CO 


T  -    "        "^  *"^  -  ^* 

(N  O  t—  O  *J  I—  O  "4"  X  '7^1  O  O  >^  Ci  -f  Ci  O  O  tC  7-1  X  ^  O  1— 


—  —  (N 


26 


INTEREST. 


Table  No.  4  gives  time  in  which  money  will  be  doubled  if 
it  is  invested  either  on  simple  or  compound  interest,  compounded 
annually. 

TABLE  No.  4. 


SIMPLE   INTEREST. 

COMPOUND  INTEREST. 

% 

Years. 

Days. 

% 

Years. 

Days. 

2 

50 

2 

35 

1 

*# 

40 

2# 

28 

30 

8 

33 

120 

3 

23 

162 

8# 

28 

206 

W 

20 

54 

4 

25 

4 

17 

240 

4/2 

22 

80 

4X 

15 

168 

5 

20 

5 

14 

75 

6 

16 

240 

6 

11 

321 

7 

14 

103 

7 

10 

89 

8 

12 

180 

8 

9 

2 

9 

11 

40 

9 

8 

16 

10 

10 

10 

7 

98 

11 

9 

33 

11 

6 

231 

12 

8 

120 

12 

6 

42 

Results  of  Saving  Small  Amounts  of  Honey. 

The  following  shows  how  easy  it  is  to  accumulate  a  fortune, 
provided  proper  steps  are  taken. 

The  table  gives  the  result  of  daily  savings,  put  in  a  savings 
bank  paying  4  per  cent,  per  year,  computed  semi-annually : 


Savings 
per  Day. 

Savings 
per  Mo. 

Amount  in 
5  years. 

Amount  in 
10  years. 

Amount  in 
15  years. 

Amount  in 
20  years. 

Amount  in 
25  years. 

.05 
.10 
.25 
.50 
.75 
$1.00 

$  1.20 
2.40 
6.00 
12.00 
18.00 
24.00 

$     78.84 
157.68 
394.20 
788.40 
1,182.60 
1,576.80 

$    174.96 
349.92 
874.80 
1,749.60 
2,624.40 
3,499.20 

$   292.07 
584.15 
1,460.37 
2,920.74 
4,381.11 
5,841.48 

$   434.88 
869.76 
2,174.40 
4,348.80 
6,523.20 
8,697.60 

$     608.94 
1,217.88 
3,044.74 

6,089.48 
9,133.22 
12,178.96 

Nearly  every  person  wastes  an  amount  in  twenty  or  thirty 
years,  which,  if  saved  and  carefully  invested,  would  make  a 
family  quite  independent;  but  the  principle  of  small  savings 
has  been  lost  sight  of  in  the  general  desire  to  become  wealthy. 


EQUATION    OF    PAYMENTS.  2  7 

EQUATION   OF   PAYMENTS. 

When  several  debts  are  due  at  different  dates  the  average 
time  when  all  the  debts  are  due  is  calculated  by  the  following 
rule: 

Multiply  each  debt  separately  by  the  number  of  days  be- 
tween its  own  date  of  maturity  and  the  date  of  the  debt  earliest 
due.  Divide  the  sum  of  these  products  by  the  sum  of  the  debts  ; 
the  quotient  will  express  the  number  of  days  subsequent  to  the 
leading  day  when  the  whole  debt  should  be  paid  in  one  sum. 

EXAMPLE. 

A  owed  to  B  the  following  sums:  $250  due  May  12,  $120 
due  July  19,  $410  due  August  1(5,  and  $60  due  September  21,  all 
in  the  same  year.  When  should  the  whole  sum  be  paid  at  once 
in  order  that  neither  shall  lose  any  interest? 

Solution  : 

May  12 $250 

May  12  to  July  11)  is  08  days ;      120  X    68  =     8160 

May  12  to  Aug.  16  is  96  days;  410  X    96  =  39360 

May  12  to  Sept.  21  is  132  days;    60  X  132  =     7920 

$840  )  55440  =   65.9 

66  days  after  May  12  will  be  July  17. 

When  several  debts  are  due  after  different  lengths  of  time, 
the  average  time  is  calculated  by  this  rule :  Multiply  the  debt 
by  the  time ;  divide  the  sum  of  the  products  by  the  sum  of  the 
debts,  and  the  quotient  is  the  time  when  all  the  debts  may  be 
considered  due. 

EXAMPLE. 

A  owed  B  $600,  due  in  7  months  ;  $200  due  in  one  month,  and 
$700  due  in  3  months.  When  should  the  whole  debt  be  paid  in 
one  sum  in  order  that  neither  shall  lose  any  interest? 

Solution : 

(500  X  7  =  4200 
700  X  3  =  2100 
200  X  1  =  200 

1500  )  6500  =  4J4  months. 

NOTE  :  If  the  debts  contain  both  dollars  and  cents  the 
cents  may,  if  such  refinement  is  required,  be  considered  as  deci- 
mal parts  of  a  dollar,  but  practically  in  such  problems  the  cents 
may  be  omitted  in  the  calculation. 


PARTNERSHIP, 

or  calculating  of  proportional  parts,  is  the  calculation  of  the 
parts  of  a  certain  quantity  in  such  a  way  that  the  ratio  between 
the  separate  parts  is  equal  to  the  ratio  of  certain  given  numbers. 


2  8  PARTNERSHIP. 

EXAMPLE  1. 

A  composition  for  welding  cast  steel  consists  of  9  parts  of 
borax  and   one  part  of  sal-ammoniac.      How  much   of  each, 
borax  and  sal-ammoniac,  must  be  taken  for  a  mixture  of  5  Ibs.? 
Solution  : 

T^  x  5  =  ±l/2  Ibs.  borax. 
TO  X  5  =  YZ  Ib.  sal-ammoniac. 
EXAMPLE  2. 

An  alloy  shall  consist  of  1(50  parts  of  copper,  15  parts  of  tin 
and  5  parts  of  zinc.    How  much  of  each  will  be  used  for  a  cast- 
ing weighing  360  Ibs.? 
Solution  : 

160  iffy  X  360  =  320  Ibs.  of  copper. 

15          .     r¥o  X  360  =  30  Ibs.  of  tin. 
5  T!  o  X  360  =  10  Ibs.  of  zinc. 

180 

EXAMPLE  3. 

Four  persons—  A,  B,  C  and  D,  are  buying  a  certain  amount, 
of  goods  together.  A's  part  is  $500,  B's,  $100,  C's,  $250,  and 
D's,  $150.  On  the  undertaking  they  are  clearing  a  net  profit 
of  $120.  How  much  of  this  is  each  to  have  ? 

Solution  : 

500  A's  Part  =  tV°A  X  120  =  $60 

100  B's      "      =  TV°o°<i  X  120  =    12 

250  C's      "      =  T2«fo°o-  X  120  -    30 

150  D's     "      =            X  120  =    18 


$1000 

EXAMPLE  4. 

Two  persons  —  A  and  B,  are  putting  money  into  business,  A, 
$2,000  and  B,  $3,000,  but  A  has  his  money  invested  in  the  busi- 
ness 2  years  and  B  2>^  years  ;  the  net  profit  of  the  undertaking 
is  $2,300.  How  much  is  each  to  have  of  the  profit  ? 

Solution: 

A,  2000  X  2      =    4000 

B,  3000  X  2^  =_  7500 

11500 

A's       Part  is       T¥$&  X  2300  —  $    800 
B's          "      "        rtV0o0<y  X  2300  =      1500 
In  cases  like  this  it  must  be  taken  into  consideration  that 
the  time  is  not  equal  ;  B  has  not  only  had  the  largest  capital  in- 
vested but  he  has  also  had  the  capital  at  work  in  the  business 


SQUARE    ROOT.  29 

the  longest  time,  namely,  2>£  years,  while  A  has  only  had  his 
capital  invested  2  years.  The  ratio  is,  therefore,  not  #2,000  to 
$3,000  but  #4,000  to  $7,500,  because  $2,000  in  2  years  is  equal 
to  $4,000  in  one  year,  and  $3,000  in  2^  years  is  equal  to  $7,500  in 
one  year. 


SQUARE  ROOT. 

When  the  square  root  is  to  be  extracted  the  number  is  di- 
vided into  periods  consisting  of  two  figures,  commencing  from 
the  extreme  right  if  the  number  has  no  decimals,  or  from  the 
decimal  point  towards  the  left  for  the  whole  numbers  and 
towards  the  right  for  the  decimals.  (If  the  last  period  of  deci- 
mals should  have  but  one  figure  then  annex  a  cipher,  so  that 
this  period  also  has  two  figures,  but  if  the  period  to  the  extreme 
left  in  the  integer  should  happen  to  have  only  one  figure  it 
makes  no  difference;  leave  it  as  it  is.)  Ascertain  the  highest 
root  of  the  first  period  and  place  it  to  the  right  of  the  number 
as  in  long  division.  Square  this  root  and  subtract  the  product 
of  this  from  the  first  period.  To  the  remainder  annex  the  next 
period  of  numbers.  Take  for  divisor  20  times  the  part  of  the 
root  already  found*  and  the  quotient  is  the  next  figure  in  the 
root,  if  the  product  of  this  figure  and  the  divisor  added  to  the 
square  of  the  figure  does  not  exceed  the  dividend.  To  the 
difference  between  this  sum  and  the  dividend  is  annexed  the 
next  period  of  numbers.  For  divisor  take  again  20  times  the 
part  of  the  root  already  found,  etc.  Continue  in  this  manner 
until  the  last  period  is  used.  If  there  is  any  remainder,  and  a 
more  exact  root  is  required,  ciphers  may  be  annexed  in  pairs  and 
the  operation  continued  until  as  many  decimals  in  the  root  are 
obtained  as  are  wanted. 


EXAMPLE  1. 

Extract  the  square  root  of  271,441. 


Solution :  \/27jl4  41 

52  =  25| 

20  X  5  =  100  )  214 
100  X  2  +  22  =   204 


20  X  52  =  1040  )  1041 
1040  X  1  +  I2  1041 


=  521 


0000 
Thus:    \/27M4i  =  521,  because  521  X  521  =  271,441. 

*  If  this  divisor  exceeds  the  dividend,  write  a  cipher  in  the  root;  annex  the  next 
period  of  numbers,  calculate  a  new  divisor,  corresponding  to  the  increased  rpot,  and 
proceed  as  explained. 


30  CUBE  ROOT. 

EXAMPLE  2. 

Extract  the  square  root  of  26.6256. 

Solution : 

\/266256  =5.16 


20  X  5  =  100  )  162 

100  X  1  +  I2  =  101 

20  X  51  =  1020  )    6156 

1020  X  6  +  6a  =    6156 

0000 


CUBE  ROOT. 

When  the  cube  root  is  to  be  extracted,  the  number  is 
divided  into  periods  consisting  of  three  figures.  Commencing 
from  the  extreme  right  if  the  number  has  no  decimals,  or  from 
the  decimal  point,  toward  the  left,  for  the  whole  number,  and 
toward  the  right  for  the  decimals.  (If  the  last  period  of  deci- 
mals should  not  have  three  figures,  then  annex  ciphers  until 
this  period  also  has  three  figures,  but  if  the  period  to  the 
extreme  left  in  the  integer  should  happen  to  consist  of  less  than 
three  figures  it  makes  no  difference ;  leave  it  as  it  is.)  Ascer- 
tain highest  cube  root  in  the  first  period  and  place  it  to  the 
right  of  the  number,  the  same  as  in  long  division.  Cube  this 
root  and  subtract  the  product  from  the  first  period.  To  the 
remainder  annex  next  period  of  numbers.  For  the  divisor  in 
this  number  take  300  times  the  square  of  the  part  of  the  root 
already  found,*  and  the  quotient  is  the  next  figure  in  the  root,  if 
the  product  of  this  figure  multiplied  by  the  divisor  and  added 
to  30  times  the  part  of  the  root  already  found,  multiplied  by  the 
square  of  this  quotient  and  added  to  the  cube  of  the  quotient, 
does  not  exceed  this  dividend.  To  the  difference  between  this 
sum  and  the  dividend  is  annexed  the  next  period  of  numbers. 
For  divisor  take  again  300  times  the  square  of  the  part  of  the 
root  already  found,  etc.  Continue  in  this  manner  until  the  last 
period  is  used.  If  there  is  any  remainder  from  last  period,  and 
a  more  exact  root  is  required,  ciphers  may  be  annexed  three  at 
a  time,  and  the  operation  continued  until  as  many  decimals  are 
obtained  in  the  root  as  are  waated. 


*  If  this  divisor  exceeds  the  dividend,  write  a  cipher  in  the  root,  annex  the  next 
period  of  numbers,  calculating  a  new  divisor  corresponding  to  the  increased  root, 
and  proceed  as  explained. 


CUBE    ROOT.  31 


EXAMPLE  l. 

Extract  the  cube  root  of  275,894,451. 

Solution : 


X/2751894 

68  =  216| 

300  X  62  ==  10800  )    59894 
10800  X  5  +  30  X  6  X  52  +  53  =    58625 


451 


300  X  652  =  1207500  )  1269451 
1267500  X  1  +  30  X  65  X  I2  +  I3  =  1269451 


651 


0000000 
Thus : 

3    

\/275,894,451  =  651,  because  651  X  651  X  651  =  275,894,451. 

EXAMPLE  2  : 

Extract  the  cube  root  of  551.368. 

Solution :  3 


\/551 
83  =  512 


8.2 


300  X  82  =  19200  )    39368 
19200  X  2  +  30  X  8  X  22  +  23  =_39368 

00000 

The  square  root  of  a  number  consisting  of  two  figures 
will  never  consist  of  more  than  one  figure,  and  the  square 
root  of  a  number  consisting  of  four  figures  will  never  consist  of 
more  than  two  figures;  hence,  the  rule  to  divide  numbers  into 
periods  consisting  of  two  figures. 

The  cube  root  of  a  number  consisting  of  three  figures  will 
never  consist  of  more  than  one  figure,  and  the  cube  root  of  a 
number  consisting  of  six  figures  will  never  consist  of  more  than 
two  figures ;  hence,  the  rule  to  divide  the  numbers  into  periods 
consisting  of  three  figures. 

There  will  always  be  one  decimal  in  the  root  for  each 
period  of  decimals  in  the  number  of  which  the  root  is  extracted. 
This  relates  to  both  cube  and  square  root. 

The  root  of  a  fraction  may  be  found  by  extracting  the 
separate  roots  of  numerator  and  denominator,  or  the  fraction 
may  be  first  reduced  to  a  decimal  fraction  before  the  root  is 
extracted. 

The  root  of  a  mixed  number  may  be  extracted  by  first  re- 
ducing the  number  to  an  improper  fraction  and  then  extracting 
the  separate  roots  of  numerator  and  denominator,  or  the  number 
may  be  first  reduced  to  consist  of  integer  and  decimal  fractions, 
and  the  root  extracted  as  usual. 


32  CUBE  ROOT. 

Radical  Quantities  Expressed  without  the  Radical  Sign. 

The  radical  sign  is  not  always  used  in  signifying  radical 
quantities.  Sometimes  a  quantity  expressing  a  root  is  written 
as  a  quantity  to  be  raised  into  a  fractional  power.  For  instance : 

>/16~may  be  written  16^.     This  is  the  same  value ;  thus, 
Vl6~=  4  and  16*  =  4. 

3 

V  27  may  be  written,  27*  =  3. 

3 3    

8t  =  V82  =  >V  64  =4. 

The  denominator  in  the  exponent  always  indicates  which 
root  is  to  be  extracted.  Thus,  8^  will  be  square  8  and  extract 
the  cube  root  from  the  product. 

EXAMPLE. 

*,_  4  __ 

16l  =  V 163  =  V  4096  =  8. 
Thus,  cube  16  and  extract  the  fourth  root  of  the  product. 


RECIPROCALS. 

The  reciprocal  of  any  number  is  the  quotient  which  is  ob- 
tained when  1  is  divided  by  the  number.  For  instance,  the 
reciprocal  of  4  is  X  —  0-25  5  the  reciprocal  of  16  is  Taff  = 
0.0625,  etc. 

Frequently  it  is  a  saving  of  time  when  performing  long 
division  to  use  the  reciprocal,  as  multiplying  the  dividend  by 
the  reciprocal  of  the  divisor  gives  the  quotient.  For  instance, 
divide  4  by  758.  In  Table  No.  6  the  reciprocal  of  758  is  given 
as  0.0013193.  Multiplying  0.0013193  by  4  gives  0.0052772, 
which  is  correct  to  six  decimals.  When  reducing  vulgar  frac- 
tions to  decimals  the  reciprocal  may  be  used  with  advantage. 
For  instance,  reduce  £j  to  decimals.  In  Table  No.  6  the  recip- 
rocal of  64  is  given  as  0.015625,  and  15  X  0.015625  =  0.234375, 
which  is  the  decimal  of  ££. 

IMPORTANT. — Whenever  the  exact  reciprocal  is  not  ex- 
pressible by  decimals  the  result  obtained  by  its  use  is,  as 
explained  above,  only  approximate. 


SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


33 


TABLE    No.  5.     Giving   Squares,  Cubes,   Square    Roots, 

Cube  Roots,  and  Reciprocals  of  Fractions  and 

flixed  Numbers,  from  J    to  10. 


n 

V 

n3 

VT 

^T 

1 

n 

A 

0.000244 

0.0000038 

0.125 

0.25 

64 

A 

0.000977 

0.0000305 

0.17678 

0.31496 

32 

A 

0.002196 

0.000103 

0.21651 

0.36056 

21.3333 

A 

0.003906 

0.000244 

0.25 

0.39685 

16 

A 

0.006104 

0.000477 

0.27951 

0.42750 

12.8 

A 

0.008789 

0.000823 

0.30619 

0.45428 

10.6667 

A 

0.011963 

0.001308 

0.33072 

0.47823 

9.1428 

i 

0.015625 

0.001953 

0.35355 

0.5 

8 

A 

0.01977 

0.00278 

0.375 

0.52002 

7.1111 

A 

0.02441 

0.00381 

0.39528 

0.53861 

6.4 

a 

0.02954 

0.00508 

0.41458 

0.55599 

5.8182 

A 

0.03516 

0.00659 

0.43301 

0.57236 

5.3333 

H 

0.04126 

0.00838 

0.45069 

0.58783 

4.9231 

A 

0.04785 

0.01047 

0.46771 

0.60254 

4.5714 

if 

0.05493 

0.01287 

0.48412 

0.61655 

4.2666 

j 

0.06250 

0.01562 

0.5 

0.62996 

4 

ti 

0.07056 

0.01874 

0.51539 

0.64282 

3.7647 

A 

0.07910 

0.02225 

0.53033 

0.65519 

3.5556 

H 

0.08813 

0.02616 

0.54482 

0.66709 

3.3684 

A 

0.09766 

0.03052 

0.55902 

0.67860 

3.2 

H 

0.10766 

0.03533 

0.57282 

0.68973 

3.0476 

H 

0.11816 

0.04062 

0.58630 

0.70051 

2.9091 

H 

0.12915 

0.04641 

0.59942 

0.71097 

2.7826 

i 

0.14062 

0.05273 

0.61237 

0.72112 

2.6667 

H 

0.15258 

0.05960 

0.625 

0.73100 

2.56 

li 

0.16504 

0.06705 

0.63738 

0.74062 

2.4615 

II 

0.17798 

0.07508 

0.64952 

0.75 

2.3703 

jV 

0.19141 

0.08374 

0.66144 

0.75915 

2.2857 

H 

0.20522 

0.09303 

0.67314 

0.76808 

2.2069 

H 

0.21973 

0.10300 

0.68465 

0.77681 

2.1333 

H 

0.23463 

0.11364 

0.69597 

0.78534 

2.0645 

i 

0.25 

0.12500 

0.70711 

0.79370 

2 

34  SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

w2 

«* 

vr 

<hr 

1 

n 

n 

H 

fi 

9 
Ttf 

0.26587 
0.28225 
0.29906 
0.31641 

0.13709 
0.14993 
0.16356 

0.17789 

0.71807 
0.72887 
0.73951 
0.75000 

0.80188 
0.80990 
0.81777 
0.82548 

1.9394 

1.8823 
1.8286 

1.7778 

If 

0.33423 

0.19315 

0.76034 

0.83306 

1.7297 

it 

0.35254 

0.20932 

0.77055 

0.84049 

1.6842 

if 

0.37134 

0.22628 

0.78062 

0.84781 

1.6410 

i 
Ii 

0.39062 
0.41040 

0.24414 
0.26291 

0.79057 
0.80039 

0.85499 
0.86205 

1.6 
1.5610 

I! 

H 

If 

0.43066 
0.45141 
0.47266 
0.49438 

0.28262 
0.30330 
0.32495 
0.34761 

0.81009 
0.81968 
0.82916 
0.83853 

0.86901 

0.87585 
0.88259 
0.88922 

1.5238 
1.4884 
1.4545 
1.4222 

I! 
i 

ii 

ft 

0.51660 
0.53931 
0.56250 

0.58618 
0.61035 

0.37131 
0.39605 
0.42187 
0.44880 
0.47684 

0.84779 
0.85696 
0.86603 
0.87500 
0.88388 

0.89576 
0.90221 
0.90856 
0.91483 
0.92101 

1.3913 
1.3617 
1.3333 
1.3061 
1.2800 

it 

it 
« 

0.63501 
0.66016 
0.68579 
0.71191 

0.50602 
0.53638 
0.56792 
0.60068 

0.89268 
0.90139 
0.91001 
0.91856 

0.92711 
0.93313 
0.93907 
0.94494 

1.2549 
1.2308 
1.2075 
1.1852 

ii 

0.73853 

0.63467 

0.92702 

0.95074 

1.1636 

« 

ft 

0.76562 
0.79321 
0.82129 

0.66992 
0.70646 
0.74429 

0.93541 
0.94373 
0.95197 

0.95647 
0.96213 
0.96772 

1.1428 
1.1228 
1.1034 

Ii 
11 

0.84985 
0.87891 

0.78346 
0.82397 

0.96014 
0.96825 

0.97325 
0.97872 

1.0847 
1.0667 

fi 
fi 
fl 

0.90845 
0.90848 
0.96899 

0.86586 
0.90915 
0.95385 

0.97628 
0.98425 
0.99216 

0.98412 
0.98947 
0.99476 

1.0492 
1.0323 
1.01587 

1 

1 

1 

1 

1 

1 

1  i 

1.12891 

1.19943 

1.03078 

1.02041 

0.94118 

4 

1.26562 

1.42323 

1.06066 

1.04004 

0.88889 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS.  35 


n 

«. 

n3 

VT 

3  

1 

n 

IT"* 

1.41016 

1.67456 

1.08965 

1.05896 

0.84211 

if 

i  A 

1.5625 
1.72266 
1.89062 

1.953125 
2.26099 
2.59961 

1.11803 
1.14564 
1.17260 

1.07722 
1.09488 
1.11199 

0.8 
0.76190 
0.72727 

187 

2.06641 

2.97046 

1.19896 

1.12859 

0.69565 

Jl 

2.25 

3.375 

1.22474 

1.14471 

0.66667 

I-9 

2.44141 

3.81470 

1.25 

1.16040 

0.64 

if 

2.640625 
2.84766 

4.29102 
4.80542 

1.27475 
1.29904 

1.17567 
1.19055 

0.61539 
0.59260 

!f 

3.0625 

5.35937 

1.32288 

1.20507 

0.57143 

Hi 

3.28516 
3.515625 

5.95434 
6.59180 

1.34630 
1.36931 

1.21925 
1.23311 

0.55172 
0.53333 

HI 

3.75391 

7.27319 

1.39194 

1.24666 

0.51613 

2 

4 

8 

1.41421 

1.25992 

0.5 

2  * 

4.25390 

8.77368 

1.43614 

1.27291 

0.48485 

2C 

4.515625 
4.78516 

9.59582 
10.46753 

1.45774 
1.47902 

1.28564 
1.29812 

0.47059 
0.45714 

2™ 

2* 
2« 

2T'ff 

H 

5.0625 
5.34766 
5.640625 
5.94141 
6.25 
6.56541 

11.390625 
12.36646 
13.39648 
14.48217 
15.625 
16.82641 

1.5 
1.52069 
1.54110 
1.56125 
1.58114 
1.60078 

1.31037 
1.32239 
1.33420 
1.34580 
1.35721 
1.36843 

0.44444 
0.43243 
0.42105 
0.41026 
0.4 
0.39024 

2| 

6.890625 

18.08789 

1.62018 

1.37946 

0.38095 

2Ji 

7.22266 

19.41090 

1.63936 

1.39032 

0.37209 

23 

7.5625 

20.79687 

1.65831 

1.40101 

0.36364 

2« 
2* 

7.91016 
8.265625 
8.62891 

22.24731 
23.76367 
25.34724 

.67705 
.69558 
.71391 

1.41155 
1.42193 
1.43216 

0.35555 
0.34783 
0.34042 

3 

9 

27 

.73205 

1.44225 

0.33333 

3j 

9.765625 

30.51758 

.76777 

1.46201 

0.32 

31 

10.5625 

34.32812 

.80278 

1.48125 

0.3077 

3I 

11.390625 

38.44336 

.83712 

1.5 

0.2963 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


W 

tf2 

W3 

VT 

3  

« 

8i 

3| 
8| 

12.25 
13.140625 
14.0625 

42.875 
47.63476 
52.73437 

1.87083 
1.90394 
1.93649 

1.51829 
1.53616 
1.55362 

0.28571 
0.27586 
0.26667 

35 

15.015625 

58.18555 

1.96850 

1.57069 

0.25806 

4 

16 

64 

2 

1.58740 

0.25 

4J 

4| 

18.0625 
20.25 
22.5625 

76.76562 
91.125 
107.17187 

2.06155 
2.12132 
2.17945 

1.61981 
1.65096 
1.68099 

0.23529 
0.22222 
0.21053 

5 

25 

125 

2.23607 

1.70998 

0.2 

5i 

27.5625 

144.70312 

2.291288 

1.73801 

0.19048 

5J 

30.25 
33.0625 

166.375 
190.10937 

2.34521 
2.39792 

1.76517 
1.79152 

0.18182 
0.17391 

6 

36 

216 

2.44949 

1.81712 

0.16667 

rH^  fHlN  tO'-l< 

CO  CO  CO 

39.0625 
42.25 
45.5625 

244.140625 
274.625 
307.54687 

2.5 
2.54951 

2.59808 

1.84202 
.86626 

.88988 

0.16 
0.15385 
0.14815 

7 

49 

343 

2.64575 

.91293 

0.14286 

7-i- 

52.5625 

381.07812 

2.69258 

.93544 

0.13793 

7-i- 

56.25 

421.875 

2.73861 

.95743 

0.13333 

7J 

60.0625 

465.48437 

2.78388 

.97895 

0.12903 

8 

64 

512 

2.82843 

2 

0.125 

8i 

68.0625 

561.5156 

2.87228 

2.02062 

0.12121 

» 

8| 

72.25 
76.5625 

614.125 
669.92187 

2.91548 
2.95804 

2.04083 
2.06064 

0.11765 
0.11428 

9 

81 

729 

3 

2.08008 

0.111111 

91 
9t 

85.5625 
90.25 
95.0625 

791.4531 

857.375 
926.8594 

3.04138 
3.08221 
3.1225 

2.09917 
2.11791 
2.13633 

0.10811 
0.10526 
0.10256 

10 

100 

1000 

3.16228 

2.15443 

0.1 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


37 


TABLE  No.  6.     Giving  Squares,  Cubes,   Square  Roots, 

Cube  Roots,  and  Reciprocals  of  Numbers  from 

o.i  to  1000. 


n 

w2 

n3 

V~n 

3 

V  n 

1 

n 

0.1 

0.01 

0.001 

0.31623 

0.46416 

10 

0.2 

0.04 

0.008 

0.44721 

0.58480 

5 

0.3 

0.09 

0.027 

0.54772 

0.66943 

3.3333 

0.4 

0.16 

0.064 

0.63245 

0.73681 

2.5 

0.5 

0.25 

0.125 

0.70711 

0.79370 

2 

0.6 

0.36 

0.216 

0.774597 

0.84343 

1.66667 

0.7 

0.49 

0.343 

0.83666 

0.88790 

1.42857 

0.8 

0.68 

0.512 

0.89443 

0.92832 

1.25 

0.9 

0.81 

0.729 

0.94868 

0.96549 

1.11111 

1 

1 

1 

1 

1 

1 

1.1 

1.21 

1.331 

1.04881 

1.03228 

0.909091 

1.2 

1.44 

1.728 

1.09545 

1.06266 

0.833333 

1.3 

1.69 

2.197 

1.14018 

1.09134 

0.769231 

A 

.  1.96 

2.744 

1.18322 

1.11869 

0.714286 

.5 

2.25 

3.375 

1.22475 

1.14471 

0.666667 

.6 

2.56 

4.096 

1.26491 

1.16961 

0.625 

.7 

2.89 

4.913 

1.30384 

1.19347 

0.588235 

.8 

3.24 

5.832 

1.34164 

1.21644 

0.555556 

1.9 

3.61 

6.859 

1.37840 

1.23855 

0.526316 

2 

4 

8 

1.41421 

.25992 

0.5 

2.1 

4.41 

9.261 

1.449138 

.28058 

0.476190 

2.2 

4.84 

10.648 

1.48324 

30059 

0.454545 

2.3 

5.29 

12.167 

1.51657 

.32001 

0.434783 

2.4 

5.76 

13.824 

1.54919 

.33887 

0.416667 

2.5 

6.25 

15.625 

1.58114 

.35721 

0.4 

2.6 

6.76 

17.576 

1.61245 

1.37508 

0.384615 

2.7 

7.29 

19.683 

1.64317 

.39248 

0.37037 

2.8 

7.84 

21.952 

1.67332 

.40946 

0.357143 

2.9 

8.41 

24.389 

1.70294 

.42604 

0.344828 

3 

9 

27 

1.73205 

1.44225 

0.333333 

3.2 

10.24 

32.768 

1.78885 

1.47361 

0.3125 

3.4 

11.56 

39.304 

1.84391 

1.50369 

0.294118 

3.6 

12.96 

46.656 

1.89737 

1.53262 

0.277778 

3.8 

14.44 

54.872 

1.94936 

1.56089 

0.263158 

4 

16 

64 

2 

1.58740 

0.25 

4.2 

17.64 

74.088 

2.04939 

1.61343 

0.238095 

4.4 

19.36 

85.184 

2.09762 

1.63868 

0.227273 

4.6 

21.16 

97.336 

2.14476 

1.66310 

0.217391 

4.8 

23.04 

110.592 

2.19089 

1.68687 

0.208333 

5 

25 

125 

2.23607 

1.70998 

0.2 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

W2 

n3 

v^ 

^ 

1 

n 

6 

36 

216 

2.44949 

1.81712 

0.1666667 

7 

49 

343 

2.64575 

1.91293 

0.142857 

8 

64 

512 

2.82843 

2 

0.125 

9 

81 

729 

3 

2.08008 

0.111111 

10 

100 

1000 

3.16228 

2.15443 

0.1 

11 

121 

1331 

3.31662 

2.22398 

0.0909091 

12 

144 

1728 

3.46410 

2.28943 

0.0833333 

13 

169 

2197 

3.60555 

2.35133 

0.0769231 

14 

196 

2744 

3.74166 

2.41014 

0.0714286 

15 

225 

3375 

3.87298 

2.46621 

0.0666667 

16 

256 

4096 

4 

2.51984 

0.0625 

17 

289 

4913 

4.12311 

2.57128 

0.0588235 

18 

324 

5832 

4.24264 

2.62074 

0.0555556 

19 

361 

6859 

4.35890 

2.66840 

0.0526316 

20 

400 

8000 

4.47214 

2.71442 

0.05 

21 

441 

9261 

4.58258 

2.75892 

0.0476190 

22 

484 

10648 

4.69042 

2.80204 

0.0454545 

23 

529 

12167 

4.79583 

2.84387 

0.0434783 

24 

576 

13824 

4.89898 

2.88450 

0.0416667 

25 

625 

15625 

5 

2.92402 

0.04 

26 

676 

17576 

5.09902 

2.96250 

0.0384615 

27 

729 

19683 

5.19615 

3 

0.0370370 

28 

784 

21952 

5.29150 

3.03659 

0.0357143 

29 

841 

24389 

5.38516 

3.07232 

0.0344828 

30 

900 

27000 

5.47723 

3.10723 

0.0333333 

31 

961 

29791 

5.56776 

3.14138 

0.0322581 

32 

1024 

32768 

5.65685 

3.17480 

0.03125 

33 

1089 

35937 

5.74456 

3.20753 

0.0303030 

34 

1156 

39304 

5.83095 

3.23961 

0.0294118 

35 

1225 

42875 

5.91608 

3.27107 

0.0285714 

36 

1296 

46656 

6 

3.30193 

0.0277778 

37 

1369 

50653 

6.08276 

3.33222 

0.0270270 

38 

1444 

54872 

6.16441 

3.36198 

0.0263158 

39 

1521 

59319 

6.245 

3.39121 

0.0256410 

40 

1600 

64000 

6.32456 

3.41995 

0.025 

41 

1681 

68921 

6.40312 

3.44822 

0.0243902 

42 

1764 

74088 

6.48074 

3.47603 

0.0238095 

43 

1849 

79507 

6.55744 

3.50340 

0.0232558 

44 

1936 

85184 

6.63325 

3.53035 

0.0227273 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


39 


n 

W2 

n3 

v7~ 

te 

1 

n 

45 

2025 

91125 

6.70820 

3.55689 

0.0222222 

46 

2116 

97336 

6.78233 

3.58305 

0.0217391 

47 

2209 

103823 

6.85565 

3.60883 

0.0212766 

48 

2304 

110592 

6.92820 

3.63424 

0.0208333 

49 

2401 

117649 

7 

3.65931 

0.0204082 

50 

2500 

125000 

7.07107 

3.68403 

0.02 

51 

2601 

132651 

7.14143 

3.70843 

0.0196078 

52 

2704 

140608 

7.21110 

3.73251 

0.0192308 

53 

2809 

148877 

7.28011 

3.75629 

0.0188679 

54 

2916 

157464 

7.34847 

3.77976 

0.0185185 

55 

3025 

166375 

7.41620 

3.80295 

0.0181818 

56 

3136 

175616 

7.48331 

3.82586 

0.0178571 

57 

3249 

185193 

7.54983 

3.84852 

0.0175439 

58 

3364 

195112 

7.61577 

3.87088 

0.0172414 

59 

3481 

205379 

7.68115 

3.89300 

0.0169492 

60 

3600 

216000 

7.74597 

3.91487 

0.0166667 

61 

3721 

226981 

7.81025 

3.93650 

0.0163934 

62 

3844 

238328 

7.87401 

3.95789 

0.0161290 

63 

3969 

250047 

7.93725 

3.97906 

0.0158730 

64 

4096 

262144 

8 

4 

0.0156250 

65 

4225 

274625 

8.06226 

4.02073 

0.0153846 

66 

4356 

287496 

8.12404 

4.04124 

0.0151515 

67 

4489 

300763 

8.18535 

4.06155 

0.0149254 

68 

4624 

314432 

8.24621 

4.08166 

0.0147059 

69 

4761 

328509 

8.30662 

4.10157 

0.0144928 

70 

4900 

343000 

8.36660 

4.12129 

0.0142857 

71 

5041 

357911 

8.42615 

4.14082 

0.0140845 

72 

5184 

373248 

8.48528 

4.16017 

0.0138889 

73 

5329 

389017 

8.54400 

4.17934 

0.0136986 

74 

5476 

405224 

8.60233 

4.19834 

0.0135135 

75 

5625 

421875  - 

8.66025 

4.21716 

0.0133333 

76 

5776 

438976 

8.71780 

4.23582 

0.0131579 

77 

5929 

456533 

8.77496 

4.25432 

0.0129870 

78 

6084 

474552 

8.83176 

4.27266 

0.0128205 

79 

6241 

493039 

8.88819 

4.29084 

0.0126582 

80 

6400 

512000 

8.94427 

4.30887 

0.0125 

81 

6561 

531441 

9 

4.32675 

0.0123457 

82 

6724 

551368 

9.05539 

4.34448 

0.0121951 

83 

6889 

571787 

9.11043 

4.36207 

0.0120482 

84 

7056 

592704 

9.16515 

4.37952 

0.0119048 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

n2 

n3 

Vn 

^ 

1 

n 

85 

7225 

614125 

9.21954 

4.39683 

0.0117647 

86 

7396 

636056 

9.27362 

4.414 

0.0116279 

87 

7569 

658503 

9.32738 

4.43105 

0.0114943 

88 

7744- 

681472 

9.38083 

4.44797 

0.0113636 

89 

7921 

704969 

9.43398 

4.46475 

0.0112360 

90 

8100 

729000 

9.48683 

4.48140 

0.0111111 

91 

8281 

753571 

9.53939 

4.49794 

0.0109890 

92 

8464 

778688 

9.59166 

4.51436 

0.0108696 

93 

8649 

804357 

9.64365 

4.53065 

0.0107527 

94 

8836 

830584 

9.69536 

4.54684 

0.0106383 

95 

9025 

857375 

9.74679 

4.56290 

0.0105263 

96 

9216 

884736 

9.79796 

4.57886 

0.0104167 

97 

9409 

912673 

9.84886 

4.59470 

0.0103093 

98 

9604 

941192 

9.89949 

4.61044 

0.0102041 

99 

9801 

970299 

9.94987 

4.62607 

0.0101010 

100 

10000 

1000000 

10 

4.64159 

0.01 

101 

10201 

1030301 

10.04988 

4.65701 

0.0099010 

102 

10404 

1061208 

10.09950 

4.67233 

0.0098039 

103 

10609 

1092727 

10.14889 

4.68755 

0.0097087 

104 

10816 

1124864 

10.19804 

4.70267 

0.0096154 

105 

11025 

1157625 

10.24695 

4.71769 

0.0095238 

106 

11236 

1191016 

10.29563 

4.73262 

0.0094340 

107 

11449 

1225043 

10.34408 

4.74746 

0.0093458 

108 

11664 

1259712 

10.39230 

4.76220 

0.0092593 

109 

11881 

1295029 

10.44031 

4.77686 

0.0091743 

110 

12100 

1331000 

10.48809 

4.79142 

0.0090909 

111 

12321 

1367631 

10.53565 

4.80590 

0.0090090 

112 

12544 

1404928 

10.58301 

4.82028 

0.0089286 

1.18 

12769 

1442897 

10.63015 

4.83459 

0.0088496 

114 

12996 

1481544 

10.67708 

4.84881 

0.0087719 

115 

13225 

1520875 

10.72381 

4.86294 

0.0086957 

116 

13456 

1560896 

10.77033 

4.877 

0.0086207 

117 

13689 

1601613 

10.81665 

4.89097 

0.0085470 

118 

13924 

1643032 

10.86278 

4.90487 

0.0084746 

119 

14161 

1685159 

10.90871 

4.91868 

0.0084034 

120 

14400 

1728000 

10.95445 

4.93242 

0.0083333 

121 

14641 

1771561 

11 

4.94609 

0.0082645 

122 

14884 

1815848 

11.04536 

4.95968 

0.0081967 

123 

15129 

1860867 

11.09054 

4.97319 

0.0081301 

124 

15376 

1906624 

11.13553 

4.98663 

0.0080645 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

H2 

n3 

VT 

3  

V  n 

1 

n 

125 

15625 

1953125 

11.18034 

5 

0.008 

126 

15876 

2000376 

11.22497 

5.01330 

0.0079365 

127 

16129 

2048383 

11.26943 

5.02653 

0.0078740 

128 

16384 

2097152 

11.31371 

5.03968 

0.0078125 

129 

16641 

2146689 

11.35782 

5.05277 

0.0077519 

130 

16900 

2197000 

11.40175 

5.06580 

0.0076923 

131 

17161 

2248091 

11.44552 

5.07875 

0.0076336 

132 

17424 

2299968 

11.48913 

5.09164 

0.0075758 

133 

17689 

2352637 

11.53256 

5.10447 

0.0075188 

134 

17956 

2406104 

11.57584 

5.11723 

0.0074627 

135 

18225 

2460375 

11.61895 

5.12993 

0.0074074 

136 

18496 

2515456 

11.66190 

5.14256 

0.0073529 

137 

18769 

2571353 

11.70470 

5.15514 

0.0072993 

138 

19044 

2628072 

11.74734 

5.16765 

0.0072464 

139 

19321 

2685619 

11.78983 

5.18010 

0.0071942 

140 

19600 

2744000 

11.83216 

5.19249 

0.0071429 

141 

19881 

2803221 

11.87434 

5.20483 

0.0070922 

142 

20164 

2863288 

11.91638 

5.21710 

0.0070423 

143 

20449 

2924207 

11.95826 

5.22932 

0.0069930 

144 

20736 

2985984 

12 

5.24148 

0.0069444 

145 

21025 

3048625 

12.04159 

5.25359 

0.0068966 

146 

21316 

3112136 

12.08305 

5.26564 

0.0068493 

147 

21609 

3176523 

12.12436 

5.27763 

0.0068027 

148 

21904 

3241792 

12.16553 

5.28957 

0.0067568 

149 

22201 

3307949 

12.20656 

5.30146 

0.0067114 

150 

22500 

3375000 

12.24745 

5.31329 

0.0066667 

151 

22801 

3442951 

12.28821 

5.32507 

0.0066225 

152 

23104 

3511808 

12.32883 

5.33680 

0.0065789 

153 

23409 

3581577 

12.36932 

5.34848 

0.0065359 

154 

23716 

3652264 

12.40967 

5.36011 

0.0064935 

155 

24025 

3723875 

12.44990 

5.37169 

0.0064516 

156 

24336 

3796416 

12.49 

5.38321 

0.0064103 

157 

24649 

3869893 

12.52996 

5.39469 

0.0063694 

158 

24964 

3944312 

12.56981 

5.40612 

0.0063291 

159 

25281 

4019679 

12.60952 

5.41750 

0.0062893 

160 

25600 

4096000 

12.64911 

5.42884 

0.00625 

161 

25921 

4173281 

12.68858 

5.44012 

0.0062112 

162 

26244 

4251528 

12.72792 

5.45136 

0.0061728 

163 

26569 

4330747 

12.76715 

5.46256 

0.0061350 

164 

26896 

4410944 

12.80625 

5.47370 

0.0060976 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

W2 

n3 

V^T 

3 

Vn 

1 

n 

165 

27225 

4492125 

12.84523 

5.48481 

0.0060606 

166 

27556 

4574296 

12.88410 

5.49586 

0.0060241 

167 

27889 

4657463 

12.92285 

5.50688 

0.0059880 

168 

28224 

4741632 

12.96148 

5.51785 

0.0059524 

169 

28561 

4826809 

13 

5.52877 

0.0059172 

170 

28900 

4913000 

13.03840 

5.53966 

0.0058824 

171 

29241 

5000211 

13.07670 

5.55050 

0.0058480 

172 

29584 

5088448 

13.11488 

5.56130 

0.0058140 

173 

29929 

5177717 

13.15295 

5.57205 

0.0057803 

174 

30276 

5268024 

13.19091 

5.58277 

0.0057471 

175 

30625 

5359375 

13.22876 

5.59344 

0.0057143 

176 

30976 

5451776 

13.26650 

5.60408 

0.0056818 

m 

31329 

5545233 

13.30413 

5.61467 

0.0056497 

178 

31684 

5639752 

13.34165 

5.62523 

0.0056180 

179 

32041 

5735339 

13.37909 

5.63574 

0.0055866 

180 

32400 

5832000 

13.41641 

5.64622 

0.0055556 

181 

32761 

5929741 

13.45362 

5.65665 

0.0055249 

182 

33124 

6028568 

13.49074 

5.66705 

0.0054945 

183 

33489 

6128487 

13.52775 

5.67741 

0.0054645 

184 

33856 

6229504 

13.56466 

5.68773 

0.0054348 

185 

34225 

6331625 

13.60147 

5.69802 

0.0054054 

186 

34596 

6434850 

13.63818 

5.70827 

0.0053763 

187 

34969 

6539203 

13.67479 

5.71848 

0.0053476 

188 

35334 

6644672 

13.71131 

5.72865 

0.0053191 

189 

35721 

6751269 

13.74773 

5.73879 

0.0052910 

190 

36100 

6859000 

13.78405 

5.74890 

0.0052632 

191 

36481 

6967871 

13.82028 

5.75897 

0.0052356 

192 

36864 

7077888 

13.85641 

5.769 

0.0052083 

193 

37249 

7189057 

13.89244 

5.779 

0.0051813 

194 

37636 

7301384 

13.92839 

5.78896 

0.005154(5 

195 

38025 

7414875 

13.96424 

5.79889 

0.0051282 

196 

38416 

7529536 

14 

5.80879 

0.0051020 

197 

38809 

7645372 

14.03567 

5.81865 

0.0050761 

198 

39204 

7762392 

14.07125 

5.82848 

0.0050505 

199 

39601 

7880599 

14.10674 

5.83827 

0.0050251 

200 

40000 

8000000 

14.14214 

5.84804 

0.005 

201 

40401 

8120601 

14.17745 

5.45777 

0.0049751 

202 

40804 

8242408 

14.21267 

5.86747 

0.0049505 

203 

•41209 

8365427 

14.24781 

5.87713 

0.0049261 

204 

41616 

8489664 

14.28286 

5.88677 

0.0049020 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


43 


n 

». 

n3 

VT 

v^ 

J_ 

n 

205 

42025 

8615125 

14.31782 

5.89637 

0.0048781 

206 

42436 

8741816 

14.35270 

5.90594 

0.0048544 

207 

42849 

8869743 

14.38749 

5.91548 

0.0048309 

208 

43264 

8998912 

14.42221 

5.92499 

0.0048077 

209 

43681 

9129320 

14.45683 

5.93447 

0.0047847 

210 

44100 

9261000 

14.49138 

5.94392 

0.0047619 

211 

44521 

9393931 

14.52584 

5.95334 

0.0047393 

212 

44944 

9528128 

14.56022 

5.96273 

0.0047170 

213 

45369 

9663597 

14.59452 

5.97209 

0.00461)48 

214 

45796 

9800344 

14.62874 

5.98142 

0.0046729 

215 

46225 

5)1)38375 

14.66288 

5.99073 

0.0046512 

218 

46656 

10077696 

14.69694 

6 

0.0046296 

217 

47089 

10218313 

14.73092 

6.00925 

0.0046083 

218| 

47524 

10360232 

14.76482 

6.01846 

0.0045872 

219 

47961 

10503459 

14.79865 

6.02765 

0.0045662 

220 

48400 

10648000 

14.83240 

6.03681 

0.0045455 

221 

48841 

10793861 

14.86607 

6.04594 

0.0045249 

222 

49284 

10941048 

14.89966 

6.05505 

0.0045045 

223 

49729 

11089567 

14.93318 

6.06413 

0.0044843 

224 

50176 

11239424 

14.96663 

6.07318 

0.0044643 

225 

50625 

11390625 

15 

6.08220 

0.0044444 

226 

51076 

11543176 

15.03330 

6.09120 

0.0044248 

227 

51529 

11697083 

15.06652 

6.10017 

0.0044053 

228 

51984 

11852352 

15.09967 

6.10911 

0.0043860 

229 

52441 

12008989 

15.13275 

6.11803 

0.0043668 

230 

52900 

12167000 

15.16575 

6.12693 

0.0043478 

231 

53361 

12326391 

15.19868 

6.13579 

0.0043290 

232 

53824 

12487168 

15.23155 

6.14463 

0.0043103 

233 

54289 

12649337 

15.26434 

6.15345 

0.0042918 

234 

54756 

12812904 

15.29706 

6.16224 

0.0042735 

235 

55225 

12977875 

15.32971 

6.17101 

0.0042553 

236 

55696 

13144256 

15.36229 

6.17975 

0.0042373 

237- 

56169 

13312053 

15.39480 

6.18846 

0.0042194 

238 

56644 

13481272 

15.42725 

6.19715 

0.0042017 

239 

57121 

13651919 

15.45962 

6.20582 

0.0041841 

240 

57600 

13824000 

15.49193 

6.21447 

0.0041667 

241 

58081 

13997521 

15.52417 

6.22308 

0.0041494 

242 

58564 

14172488 

15.55635 

6.23168 

0.0041322 

243 

59049 

14348907 

15.58846 

6.24025 

0.0041152 

244 

59536 

14526784 

15.62050 

6.24880 

0.0040984 

44 


SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

n2 

W3 

VT 

^T 

1 

n 

245 

60025 

14706125 

15.65248 

6.25732 

0.0040816 

246 

60516 

14886936 

15.68439 

6.26583 

0.0040650 

247 

61009 

15069223 

15.71623 

6.27431 

0.0040486 

248 

61504 

15252992 

15.74802 

6.28276 

0.0040323 

249 

62001 

15438249 

15.77973 

6.29119 

0.0040161 

250 

62500 

15625000 

15.81139 

6.29961 

0.004 

251 

63001 

15813251 

15.84298 

6.30799 

0.0039841 

252 

63504 

16003008 

15.87451 

6.31636 

0.0039683 

253 

64009 

16194277 

15.90597 

6.32470 

0.0039526 

254 

64516 

16387064 

15.93738 

6.33303 

0.0039370 

255 

65025 

16581375 

15.96872 

6.34133 

0.0039216 

256 

65536 

16777216 

16 

6.34960 

0.0039062 

257 

66049 

16974593 

16.03122 

6.35786 

0.0038911 

258 

66564 

17173512 

16.06238 

6.36610 

0.0038760 

259 

67081 

17373979 

16.09348 

6.37431 

0.0038610 

260 

67600 

17576000 

16.12452 

6.38250 

0.0038462 

261 

68121 

17779581 

16.15549 

6.39068 

0.0038314 

262 

68644 

17984728 

16.18641 

6.39883 

0.0038168 

263 

69169 

18191447 

16.21727 

6.40696 

0.0038023 

264 

69696 

18399744 

16.24808 

6.41507 

0.0037879 

265 

70225 

18609625 

16.27882 

6.42316 

0.0037736 

266 

70756 

18821096 

16.30951 

6.43123 

0.0037594 

267 

71289 

19034163 

16.34013 

6.43928 

0.0037453 

268 

71824 

19248832 

16.37071 

6.44731 

0.0037313 

269 

72361 

19465109 

16.40122 

6.45531 

0.0037175 

270 

72900 

19683000 

16.43168 

6.46330 

0.0037037 

271 

73441 

19902511 

16.46208 

6.47127 

0.00369 

272 

73984 

20123648 

16.49242 

6.47922 

0.0036765 

273 

74529 

20346417 

16.52271 

6.48715 

0.0036630 

274 

75076 

20570824 

16.55295 

6.49507 

0.0036496 

275 

75625 

20796875 

16.58312 

6.50296 

0.0036364 

276 

76176 

21024576 

16.61325 

6.51083 

0.0036232 

277 

76729 

21253933 

16.64332 

6.51868 

0.0036101 

278 

77284 

21484952 

16.67333 

6.52652 

0.0035971 

279 

77841 

21717639 

16.70329 

6.53434 

0.0035842 

280 

78400 

21952000 

16.73320 

6.54213 

0.0035714 

281 

78961 

22188041 

16.76305 

6.54991 

0.0035587 

282 

79524 

22425768 

16.79286 

6.55767 

0.0035461 

283 

80089 

22665187 

16.82260 

6.56541 

0.0035336 

284 

80656 

22906304 

16.85230 

6.57314 

0.0035211 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


45 


n 

n2 

tf3 

V  n 

3  

v  n 

1 

n 

285 

81225 

23149125 

16.88194 

6.58084 

0.0035088 

286 

81796 

23393656 

16.91153 

6.58853 

0.0034965 

287 

82369 

23639903 

16.94107 

6.59620 

0.0034843 

288 

82944 

23887872 

16.97056 

6.60385 

0.0034722 

289 

83521 

24137569 

17 

6.61149 

0.0034602 

290 

84100 

24389000 

17.02939 

6.61911 

0.0034483 

291 

84681 

24642171 

17.05872 

6.62671 

0.0034364 

292 

85264 

24897088 

17.08801 

6.63429 

0.0034247 

293 

85849 

25153757 

17.11724 

6.64185 

0.0034130 

294 

86436 

25412184 

17.14643 

6.64940 

0.0034014 

295 

87025 

25672375 

17.17556 

6.65693 

0.0033898 

296 

87616 

25934336 

17.20465 

6.66444 

0.0033784 

297 

88209 

26198073 

17.23369 

6.67194 

0.0033670 

298 

88804 

26463592 

17.26268 

6.67942 

0.0033557 

299 

89401 

26730899 

17.29162 

6.68688 

0.0033445 

300 

90000 

27000000 

17.32051 

6.69433 

0.0033333 

301 

90601 

27270901 

17.34935 

6.70176 

0.0033223 

302 

91204 

27543608 

17.37815 

6.70917 

0.0033113 

303 

91809 

27818127 

17.40690 

6.71657 

0.0033003 

304 

92416 

28094464 

17.43560 

6.72395 

0.0032895 

305 

93025 

28372625 

17.46425 

6.73132 

0.0032787 

306 

93636 

28652616 

17.49286 

6.73866 

0.0032680 

307 

94249 

28934443 

17.52142 

6.746 

0.0032573 

308 

94864 

29218112 

17.54993 

6.75331 

0.0032468 

309 

95481 

29503629 

17.57840 

6.76061 

0.0032362 

310 

96100 

29791000 

17.60682 

6.76790 

0.0032258 

311 

96721 

30080231 

17.63519 

6.77517 

0.0032154 

312 

97344 

30371328 

17.66352 

6.78242 

0.0032051 

313 

97969 

30664297 

17.69181 

6.78966 

0.0031949 

314 

98596 

30959144 

17.72005 

6.79688 

0.0031847 

315 

99225 

31255875 

17.74824 

6.80409 

0.0031746 

316 

99856 

31554496 

17.77639 

6.81128 

0.0031646 

317 

100489 

31855013 

17.80449 

6.81846 

0.0031546 

318 

101124 

32157432 

17.83255 

6.82562 

0.0031447 

319 

101761 

32461759 

17.86057 

6.83277 

0.0031348 

320 

102400 

32768000 

17.88854 

6.83990 

0.0031250 

321 

103041 

33076161 

17.91647 

6.84702 

0.0031153 

322 

103684 

33386248 

17.94436 

6.85412 

0.0031056 

323 

104329 

33698267 

17.97220 

6.86121 

0.0030960 

324 

104976 

34012224 

18 

6.86829 

0.0030864 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

W2 

H3 

VrT 

<&? 

1 

n 

325 

105625 

34328125 

18.02776 

6.87534 

0.0030769 

326 

106276 

34645976 

18.05547 

6.88239 

0.0030675 

327 

106929 

34965783 

18.08314 

6.88942 

0.0030581 

328 

107584 

35287552 

18.11077 

6.89643 

0.0030488 

329 

108241 

35611289 

18.13836 

6.90344 

0.0030395 

330 

108900 

35937000 

18.16590 

6.91042 

0.0030303 

331 

109561 

36264691 

18.19341 

6.91740 

0.0030211 

332 

110224 

36594368 

18.22087 

6.92436 

0.0030120 

333 

110889 

36926037 

18.24829 

6.93131 

0.0030030 

334 

111556 

37259704 

18.27567 

6.93823 

0.0029940 

335 

112225 

37595375 

18.30301 

6.94515 

0.0029851 

336 

112896 

37933056 

18.33030 

6.95205 

0.0029762 

337 

113569 

38272753 

18.35756 

6.95894 

0.0029674 

338 

114244 

38614472 

18.38478 

6.96582 

0.0029586 

339 

114921 

38958219  ' 

18.41195 

6.97268 

0.0029499 

340 

115600 

39304000 

18.43909 

6.97953 

0.0029412 

341 

116281 

39651821 

18.46619 

6.98637 

0.0029326 

342 

116964 

40001688 

18.49324 

6.99319 

0.0029240 

343 

117649 

40353607 

18.52026 

7 

0.0029155 

344 

118336 

40707584 

18.54724 

7.00680 

0.0029070 

345 

119025 

41063625 

18.57418 

7.01358 

0.0028986 

346 

119716 

41421736 

18.60108 

7.02035 

0.0028902 

347 

120409 

41781923 

18.62794 

7.02711 

0.0028818 

348 

121104 

42144192 

18.65476 

7.03385 

0.0028736 

349 

121801 

42508549 

18.68154 

7.04059 

0.0028653 

350 

122500 

42875000 

18.70829 

7.04730 

0.0028571 

351 

123201 

43243551 

18.73499 

7.054 

0.0028490 

352 

123904 

43614208 

18.76166 

7.06070 

0.0028409 

353 

124609 

43986977 

18.78829 

7.06738 

0.0028329 

354 

125316 

44361864 

18.81489    7.07404 

0.0028249 

355 

126025 

44738875 

18.84144 

7.08070 

0.0028169 

356 

126736 

45118016 

18.86796 

7.08734 

0.0028090 

357 

127449 

45499293 

18.89444  !  7.09397 

0.0028011 

358 

128164 

45882712 

18.92089 

7.10059 

0.0027933 

359 

128881 

46268279 

18.94730 

7.10719 

0.0027855 

360 

129600 

46656000 

18.97367 

7.11379 

0.0027778 

361 

130321 

47045881 

,19 

7.12037 

0.0027701 

362 

131044 

47437928 

19.02630 

7.12694 

0.0027624 

363 

131769 

47832147 

19.05256 

7.13349 

0.0027548 

364 

132496 

48228544 

19.07878 

7.14004 

0.0027473 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


47 


n 

W2 

tt3 

w 

^ 

1 

n 

365 

133225 

48627125 

19.10497 

7.14657 

0.0027397 

366 

133956 

49027896 

19.13113 

7.15309 

0.0027322 

367 

134689 

49430863 

19.15724 

7.15960 

0.0027248 

368 

135424 

49836032 

19.18333 

7.16610 

0.0027174 

369 

136161 

50243409 

19.20937 

7.17258 

0.0027100 

370 

136900 

50(553000 

19.23538 

7.17905 

0.0027027 

371 

137641 

51064811 

19.26136 

7.18552 

0.0026954 

372 

138384 

51478848 

19.28730 

7.19197 

0.0026882 

373 

139129 

51895117 

19.31321 

7.19841 

0.0026810 

374 

139876 

52313624 

19.33908 

7.20483 

0.0026738 

375 

140625 

52734375 

19.36492 

7.21125 

0.0026667 

376 

141376 

53157376 

19.39072 

7.21765 

0.0026596 

377 

142129 

53582633 

19.41649 

7.22405 

0.0026525 

378 

142884 

54010152 

19.44222 

7.23643 

0.0026455 

379 

143641 

54439939 

19.46792 

7.23680 

0.0026385 

380 

144400 

54872000 

19.49359 

7.24316 

0.0026316 

381 

145161 

55306341 

19.51922 

7.24950 

0.0026247 

382 

145924 

55742968 

19.54482 

7.25584 

0.0026178 

383 

146689 

56181887 

19.57039 

7.26217 

0.0026110 

384 

147456 

56623104 

19.59592 

7.26848 

0.0026042 

385 

148225 

57066625 

19.62142 

7.27479 

0.0025974 

386 

148996 

57512456 

19.64688 

'  7.28108 

0.0025907 

387 

149769   57960603 

19.67232 

7.28736 

0.0025840 

388 

150544  |  58411072 

19.69772 

7.29363 

0.0025773 

389 

151321 

58863869 

19.72308 

7.29989 

0.0025707 

390 

152100 

59319000 

19.74842 

7.30614 

0.0025641 

391 

152881 

59776471 

19.77372 

7.31238 

0.0025575 

392 

153664 

60236288 

19.79899 

7.31861 

0.0025510 

393 

154449 

60698457 

19.82423 

7.32483 

0.0025445 

394 

155236 

61162984 

19.84943 

7.33104 

0.0025381 

395 

156025 

61629875 

19.87461 

7.33723 

0.0025316 

396 

156816 

62099136 

19.89975 

7.34342 

0.0025253 

397 

157609 

62570773 

19.92486 

7.34960 

0.0025189 

398 

158404 

63044792 

19.94994 

7.35576 

0.0025126 

399 

159201 

63521199 

19.97498 

7.36192 

0.0025063 

400 

160000 

64000000 

20 

7.36806 

0.0025 

401 

160801 

64481201 

20.02498 

7.37420 

0.0024938 

402 

161604 

64964808 

20.04994 

7.38032 

0.0024876 

403 

162409 

65450827 

20.07486 

7.38644 

0.0024814 

404 

163216 

65939264 

20.09975 

7.39254 

0.0024752 

48 


SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

n2 

tf3 

*/  n 

3, 

VTT 

1 

n 

405 

164025 

66430125 

20.12461 

7.39864 

0.0024691 

406 

164836 

66923416 

20.14944 

7.40472 

0.0024631 

407 

165649 

67419143 

20.17424 

7.41080 

0.0024570 

408 

166464 

67917312 

20.19901 

7.41686 

0.0024510 

409 

167281 

68417929 

20.22375 

7.42291 

0.0024450 

410 

168100 

68921000 

20.24846 

7.42896 

0.0024390 

411 

168921 

6942653  L 

20.27313 

7.43499 

0.0024331 

412 

169744 

69934528 

20.29778 

7.44102 

0.0024272 

413 

170569 

70444997 

20.32240 

7.44703 

0.0024213 

414 

171396 

70957944 

20.34699 

7.45304 

0.0024155 

415 

172225 

71473375 

20.37155 

7.45904 

0.0024096 

416 

173056 

71991296 

20.39608 

7.46502 

0.0024038 

417 

173889 

72511713 

20.42058 

7.471 

0.0023981 

418 

174724 

73034632 

20.44505 

7.47697 

0.0023923 

419 

175561 

73560059 

20.46949 

7.48292 

0.0023866 

420 

176400 

74088000 

20.49390 

7.48887 

0.0023810 

421 

177241 

74618461 

20.51828 

7.49481 

0.0023753 

422 

178084 

75151448 

20.54264 

7.50074 

0.0023697 

423 

178929 

75686967 

20.56696 

7.50666 

0.0023641 

424 

179776 

76225024 

20.59126 

7.51257 

0.0023585 

425 

180625 

76765625 

20.61553 

7.51847 

0.0023529 

426 

181476 

77308776 

20.63977 

7.52437 

0.0023474 

427 

182329 

77854483 

20.66398 

7.53025 

0.0023419 

428 

183184 

78402752 

20.68816 

7.53612 

0.0023364 

429 

184041 

78953589 

20.71232 

7.54199 

0.0023310 

430 

184900 

79507000 

20.73644 

7.54784 

0.0023256 

431 

185761 

80062991 

20.76054 

7.55369 

0.0023202 

432 

186624 

80621568 

20.78461 

7.55953 

0.0023148 

433 

187489 

81182737 

20.80865 

7.56535 

0.0023095 

434 

188356 

81746504 

20.83267 

7.57117 

0.0023041 

435 

189225 

82312875 

20.85665 

7.57698 

0.0022989 

436 

190096 

82881856 

20.88061 

7.58279 

0.0022936 

437 

190969 

83453453 

20.90455 

7.58858 

0.0022883 

438 

191844 

84027672 

20.92845 

7.59436 

0.0022831 

439 

192721 

84604519 

20.95233 

7.60014 

0.0022779 

440 

193600 

85184000 

20.97618 

7.60590 

0.0022727 

441 

194481 

85766121 

21 

7.61166 

0.0022676 

442 

195304 

86350888 

21.02380 

7.61741 

0.0022624 

443 

196249 

86938307 

21.04757 

7.62315 

0.0022573 

444 

197136 

87528384 

21.07131 

7.62888 

0.0022523 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


49 


n 

n2 

V  n 

3 

v  n 

1 

n 

445 

198025 

88121125 

21.09502 

7.63461 

0.0022472 

446 

198916 

88716536 

21.11871 

7.64032 

0.0022422 

447 

199809 

89314623 

21.14237 

7.64603 

0.0022371 

448 

200704 

89915392 

21.16601 

7.65172 

0.0022321 

449 

201601 

90518849 

21.18962 

7.65741 

0.0022272 

450 

202500 

91125000 

21.21320 

7.66309 

0.0022222 

451 

203401 

91733851 

21.23676 

7.66877 

0.0022173 

452 

204304 

92345408 

21.26029 

7.67443 

0.0022124 

453 

205209 

92959677 

21.28380 

7.68009 

0.0022075 

454 

206116 

93576664 

21.30728 

'7.68573 

0.0022026 

455 

207025 

94196375 

21.33073 

7.69137 

0.0021978 

456 

207936 

94818816 

21.35416 

7.69700 

0.0021930 

457 

208849 

95443993 

21.37756 

7.70262 

0.0021882 

458 

209764 

96071912 

21.40093 

7.70824 

0.0021834 

459 

210681 

96702579 

21.42429 

7.71384 

0.0021786 

460 

211600 

97336000 

21.44761 

7.71944 

0.0021739 

461 

212521 

97972181 

21.47091 

7.72503 

0.0021692 

462 

213444 

98611128 

21.49419 

7.73061 

0.0021645 

463 

214369 

99252847 

21.51743 

7.73619 

0.0021598 

464 

215296 

99897344 

21.54066 

7.74175 

0.0021552 

465 

216225 

100544625 

21.56386 

7.74731 

0.0021505 

466 

217156 

101194696 

21.58703 

7.75286 

0.0021459 

467 

218089 

101847563 

21.61018 

7.75840 

0.0021413 

468 

219024 

102503232 

21.63331 

7.76394 

0.0021368 

469 

219961 

103161709 

21.65641 

7.76946 

0.0021322 

470 

220900 

103823000 

21.67948 

7.77498 

0.0021277 

471 

221841 

104487111 

21.70253 

7.78049 

0.0021231 

472 

222784 

105154048 

21.72556 

7.78599 

0.0021186 

473 

223729 

105823817 

21.74856 

7.79149 

0.0021142 

474 

224676 

106496424 

21.77154 

7.79697 

0.0021097 

475 

225625 

107171875 

21.79449 

7.80245 

0.0021053 

476 

226576 

107850176 

21.81742 

7.80793 

0.0021008 

477 

227529 

108531333 

21.84033 

7.81339 

0.0020965 

478 

228484 

109215352 

21.86321 

7.81885 

0.0020921 

479 

229441 

109902239 

21.88607 

7.82429 

0.0020877 

480 

230400 

110592000 

21.90890 

7.82974 

0.0020833 

481 

231361 

111284641 

21.93171 

7.83517 

0.0020790 

482 

232324 

111980168 

21.95450 

7.84059 

0.0020747 

483 

233289 

112678587 

21.97726 

7.84601 

0.0020704 

484 

234256 

113379904 

22 

7.85142 

0.0020661 

5° 


SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

W2 

n3 

V  n 

^F 

1 

n 

485 

235225 

114084125 

22.02272 

7.85683 

0.0020619 

486 

236196 

114791256 

22.04541 

7.86222 

0.0020576 

487 

237169 

115501303 

22.06808 

7.86761 

0.0020534 

488 

238144 

116214272 

22.09072 

7.8729!) 

0.0020492 

489 

239121 

116930169 

22.11334 

7.87837 

0.0020450 

490 

240100 

117649000 

22.13594 

7.88374 

0.0020408 

491 

241081 

118370771 

22.15852 

7.88909 

0.0020367 

492 

242064 

119095488 

22.18107 

7.89445 

0.0020325 

493 

243049 

119823157 

22,20360 

7.89979 

0.0020284 

494 

244036 

120553784 

22.22611 

7.90513 

0.0020243 

495 

245025 

121287375 

22.24860 

7.91046 

0.0020202 

496 

246016 

122023936 

22.27106 

7.91578 

0.0020161 

497 

247009 

122763473 

22.29350 

7.92110 

0.0020121 

498 

248004 

123505992 

22.31591 

7.92641 

0.0020080 

499 

249001 

124251499 

22.33831 

7.93171 

0.0020040 

500 

250000 

125000000 

22.36068 

7.93701 

0.002 

501 

251001 

125751501 

22.38303 

7.94229 

0.0019960 

502 

252004 

126506008 

22.40536 

7.94757 

0.0019920 

503 

253009 

127263527 

22.42766 

7.95285 

0.0019881 

504 

254016 

128024064 

22.44994 

7.95811 

0.0019841 

505 

255025 

128787626 

22.47221 

7.96337 

0.0019802 

506 

256036 

129554216 

22.49444 

7.96863 

0.0019763 

507 

257049 

130323843 

22.51666 

7.97387 

0.0019724 

508 

258064 

131096512 

22.53886 

7.97911 

0.0019685 

509 

259081 

131872229 

22.56103 

7.98434 

0.0019646 

510 

260100 

132651000 

22.58318 

7.98957 

0.0019608 

511 

261121 

133432831 

22.60531 

7.99479 

0.0019569 

512 

262144 

134217728 

22.62742 

8 

0.0019531 

513 

263169 

135005697 

22.64950 

8.00520 

0.0019493 

514 

264196 

135796744 

22.67157 

8.01040 

0.0019455 

515 

265225 

136590875 

22.69361 

8.01559 

0.0019417 

516 

266256 

137388096 

22.71563 

8.02078 

0.0019380 

517 

267289 

138188413 

22.73763 

8.02596 

0.0019342 

518 

268324 

138991832 

22.75961 

8.03113 

0.0019205 

519 

269361 

139798359 

22.78157 

8.03629 

0.0019268 

520 

270400 

140608000 

22.80351 

8.04145 

0.0019231 

521 

271441 

141420761 

22.82542 

8.04660 

0.0019194 

522 

272484 

142236648 

22.84732 

8.05175 

0.0019157 

523 

273529 

143055667 

22.86919 

8.05689 

0.0019120 

524 

274576 

143877824 

22.89105 

8.06202 

0.0019084 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

n2 

tt3 

*S  n 

^ 

1 

n 

525 

275625 

144703125 

22.91288 

8.06714 

0.0019048 

526 

276676 

145531576!  22.93469 

8.07226 

0.0019011 

527 

277729 

146363183 

22.95648 

8.07737 

0.0018975 

528 

278784 

147197952 

22.97825 

8.08248 

0.0018939 

529 

279841 

148035889 

23 

8.08758 

0.0018904 

530 

280900 

148877000 

23.02173 

8.09267 

0.0018868 

531 

281961 

149721291 

23.04344 

8.09776 

0.0018832 

532 

283024 

150568768 

23.06513 

8.10284 

0.0018797 

533 

284089 

151419437 

23.08679 

8.10791 

0.0018762 

534 

285156 

152273304 

23.10844 

8.11298 

0.0018727 

535 

286225 

153130375 

23.13007 

8.11804 

0.0018692 

536 

287296 

153990656 

23.15167 

8.12310 

0.0018657 

537 

288369 

154854153 

23.17326 

8.12814 

0.0018622 

538 

289444 

155720872 

23.19483 

8.13319 

0.0018587 

539 

290521 

156590819 

23.21637 

8.13822 

0.0018553 

540 

291600 

157464000 

23.23790 

8.14325 

0.0018519 

541 

292681 

158340421 

23.25941 

8.14828 

0.0018484 

542 

293764 

159220088 

23.28089 

8.15329 

0.0018450 

543 

294849 

160103007 

23.30236 

8.15831 

0.0018416 

544 

295936 

160989184  i  23.32381 

8.16331 

0.0018382 

545 

297025 

161878625'  23.34524 

8.16831 

0.0018349 

546 

298116 

162771336  23.36664 

8.17330 

0.0018315 

547 

299209  !  163667323  23.38803 

8.17829 

0.0018282 

548 

300304 

164566592 

23.40940 

8.18327 

0.0018248 

549 

301401 

165469149 

23.43075 

8.18824 

0.0018215 

550 

302500 

166375000 

23.45208 

8.19321 

0.0018182 

551 

303601 

167284151 

23.47339 

8.19818 

0.0018149 

552 

304704 

168196608 

23.49468 

8.20313 

0.0018116 

553 

305809 

169112377 

23.51595 

8.20808 

0.0018083 

554 

306916 

170031464 

23.53720 

8.21303 

0.0018051 

555 

308025 

170953875 

23.55844 

8.21797 

0.0018018 

556 

309136 

171879616 

23.57965 

8.22290 

0.0017986 

557 

310249 

172808693 

23.60085 

8.22783 

0.0017953 

558 

311364 

173741112 

23.62202 

8.23275 

0.0017921 

559 

312481 

174676879 

23.64318 

8.23766 

0.0017889 

560 

313600 

175616000 

23.66432 

8.24257 

0.0017857 

561 

314721 

I765r>s4si 

23.68544 

8.24747 

0.0017825 

562 

315844 

177504328 

23.70654 

8.25237 

0.0017794 

563 

316969 

178453547 

23.72762 

8.25726 

0.0017762 

564 

318096 

179406144 

23.74868 

8.26215 

0.0017730 

52  SQUARES,    CUBES,    ROOTS,   AND    RECIPROCALS. 


n 

W2 

n3 

VT 

•fr 

1 

n 

565 

319225 

180362125 

23.76973 

8.26703 

0.0017699 

566 

320356 

181321496 

23.79075 

8.27190 

0.0017668 

567 

321489 

182284263 

23.81176 

8.27677 

0.0017637 

568 

322624 

183250432 

23.83275 

8.28163 

0.0017606 

569 

323761 

184220009 

23.85372 

8.28649 

0.0017575 

570 

324900 

185193000 

23.87467 

8.29134 

0.0017544 

571 

326041 

186169411 

23.89561 

8.29619 

0.0017513 

572 

327184 

187149248 

23.91652 

8.30103 

0.0017483 

573 

328329 

188132517 

23.93742 

8.30587 

0.0017452 

574 

329476 

189119224 

23.95830 

8.31069 

0.0017422 

575 

330625 

190109375 

23.97916 

8.31552 

0.0017391 

576 

331776 

191102976 

24 

8.32034 

0.0017361 

577 

332929 

192100033 

24.02082 

8.32515 

0.0017331 

578 

334084 

193100552 

24.04163 

8.32995 

0.0017301 

579 

335241 

194104539 

24.06242 

8.33476 

0.0017271 

580 

336400 

195112000 

24.08319 

8.33955 

0.0017241 

581 

337561 

196122941 

24.10394 

8.34434 

0.0017212 

582 

338724 

197137368 

24.12468 

8.34913 

0.0017182 

583 

339889 

198155287 

24.14539 

8.35390 

0.0017153 

584 

341056 

199176704 

24.16609 

8.35868 

0.0017123 

585 

342225 

200201625 

24.18677 

8.36345 

0.0017094 

586 

343396 

201230056 

24.20744 

8.36821 

0.0017065 

587 

344569 

202262003 

24.22808 

8.37297 

0.0017036 

588 

345744 

203297472 

24.24871 

8.37772 

0.0017007 

589 

346921 

204336469 

24.26932 

8.38247 

0.0016978 

590 

348100 

205379000 

24.28992 

8.38721 

0.0016949 

591 

349281 

206425071 

24.31049 

8.39194 

0.0016920 

592 

350464 

207474688 

24.33105 

8.39667 

0.0016892 

593 

351649 

208527857 

24.35159 

8.40140 

0.0016863 

594 

352836 

209584584 

24.37212 

8.40612 

0.0016835 

595 

354025 

210644875 

24.39262 

8.41083 

0.0016807 

596 

355216 

211708736 

24.41311 

8.41554 

0.0016779 

597 

356409 

212776173 

24.43358 

8.42025 

0.0016750 

598 

357604 

213847192 

24.45404 

8.42494 

0.0016722 

599 

358801 

214921799 

24.47448 

8.42964 

0.0016694 

600 

360000 

216000000 

24.49490 

8.43433 

0.0016667 

601 

361201 

217081801 

24.51530 

8.43901 

0.0016639 

602 

362404 

218167208 

24.53569 

8.44369 

0.0016611 

603 

363609 

219256227 

24.55606 

8.44836 

0.0016584 

604 

364816 

220348864 

24.57641 

8.45303 

0.0016556 

SQUARES,    CUBES,    ROOTS,    AND   RECIPROCALS. 


53 


n 

n2 

n3 

*/n 

3 

v  n 

1 

n 

605 

366025 

221445125 

24.59675 

8.45769 

0.0016529 

606 

367236 

222545016 

24.61707 

8.46235 

0.0016502 

607 

368449 

223648543 

24.63737 

8.46700 

0.0016474 

608 

369664 

224755712 

24.65766 

8.47165 

0.0016447 

609 

370881 

225866529 

24.67793 

8.47629 

0.0016420 

610 

372100 

226981000 

24.69818 

8.48093 

0.0016393 

611 

373321 

228099131 

24.71841 

8.48556 

0.0016367 

612 

374544 

229220928 

24.73863 

8.49018 

0.0016340 

613 

375769 

230346397 

24.75884 

8.49481 

0.0016313 

614 

376996 

231475544 

24.77902 

8.49942 

0.0016287 

615 

378225 

232608375 

24.79919 

8.50404 

0.0016260 

616 

379456 

233744896 

24.81935 

8.50864 

0.0016234 

617 

380689 

234885113 

24.83948 

8.51324 

0.0016207 

618 

381924 

236029032 

24.85961 

8.51784 

0.0016181 

619 

383161 

237176659 

24.87971 

8.52243 

0.0016155 

620 

384400 

238328000 

24.89980 

8.52702 

0.0016129 

621 

385641 

239483061 

24.91987 

8.53160 

0.0016103 

622 

386884 

240641848 

24.93993 

8.53618 

0.0016077 

623 

388129 

241804367 

24.95997 

8.54075 

0.0016051 

624 

389376 

242970624 

24.97999 

8.54532 

0.0016026 

625 

390625 

244140625 

25 

8.54988 

0.0016000 

626 

391876 

245314376 

25.01999 

8.55444 

0.0015974 

627 

393129 

246491883 

25.03997 

8.55899 

0.0015949 

628 

394384 

247673152 

25.05993 

8.56354 

0.0015924 

629 

395641 

248858189 

25.07987 

8.56808 

0.0015898 

630 

396900 

250047000 

25.09980 

8.57262 

0.0015873 

631 

398161 

251239591 

25.11971 

8.57715 

0.0015848 

632 

399424 

252435968 

25.13961 

8.58168 

0.0015823 

633 

400689 

253636137 

25.15949 

8.58622 

0.0015798 

634 

401956 

254840104 

25.17936 

8.59072 

0.0015773 

635 

403225 

256047875 

25.19921 

8.59524 

0.0015748 

636 

404496 

257259456 

25.21904 

8.59975 

0.0015723 

637 

405769 

258474853 

25.23886 

8.60425 

0.0015699 

638 

407044 

259694072 

25.25866 

8.60875 

0.0015674 

639 

408321 

260917119 

25.27845 

8.61325 

0.0015649 

640 

409600 

262144000 

25.29822 

8.61774 

0.0015625 

641 

410881 

263374721 

25.31798 

8.62222 

0.0015601 

642 

412164 

264609288 

25.33772 

8.62671 

0.0015576 

643 

413449 

265847707 

25.35744 

8.63118 

0.0015552 

644 

414736 

267089984 

25.37716 

8.63566 

0.0015528 

54 


SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

ft2 

n3 

VrT 

vn 

1 

n 

645 

416025 

268336125 

25.39685 

8.64012 

0.0015504 

646 

417316 

269586136 

25.41653 

8.64459 

0.0015480 

647 

418609 

270840023 

25.43619 

8.64904 

0.0015456 

648 

419904 

272097792 

25.45584 

8.65350 

0.0015432 

649 

421201 

273359449 

35.47548 

8.65795 

0.0015408 

650 

422500 

274625000 

25.49510 

8.66239 

0.0015385 

651 

423801 

275894451 

25.51470 

8.66683 

0.0015361 

652 

425104 

277167808 

25.53429 

8.67127 

0.0015337 

653 

426409 

278445077 

25.55386 

8.67570 

0.0015314 

654 

427716 

279726264 

25.57342 

8.68012 

0.0015291 

655 

429025 

281011375 

25.59297 

8.68455 

0.0015267 

656 

430336 

282300416 

25.61250 

8.68896 

0.0015244 

657 

431649 

283593393 

25.63201 

8.69338 

0.0015221 

658 

432964 

284890312 

25.65151 

8.69778 

0.0015198 

659 

434281 

286191179 

25.67100 

8.70219 

0.0015175 

660 

435600 

287496000 

25.69047 

8.70659 

0  0015152 

661 

436921 

288804781 

25.70992 

8.71098 

0.0015129 

662 

438244 

290117528 

25.72936 

8.71537 

0.0015106 

663 

439569 

291434247 

25.74879 

8.71976 

0.0015083 

664 

440896 

292754944 

25.76820 

8.72414 

0.0015060 

665 

442225 

294079625 

25.  78749 

8.72852 

0.0015038 

666 

443556 

295408296 

25.80698 

8.73289 

0.0015015 

667 

444889 

296740963 

25.82634 

8.73726 

0.0014993 

668 

446224 

298077632 

25.84570 

8.74162 

0.0014970 

669 

447561 

299418309 

25.86503 

8.74598 

0.0014948 

670 

448900 

300763000 

25.88436 

8.75034 

0.0014925 

671 

450241 

302111711 

25.90367 

8.75469 

0.0014903 

672 

451584 

303464448 

25.92296 

8.75904 

0.0014881 

673 

452929 

304821217 

25.94224 

8.76338 

0.0014859 

674 

454276 

306182024 

25.96151 

8.76772 

0.0014837 

675 

455625 

307546875 

25.98076 

8.77205 

0.0014815 

676 

456976 

308915776 

26 

8.77638 

0.0014793 

677 

458329 

310288733 

26.01922 

8.78071 

0.0014771 

678 

459684 

311665752 

26.03843 

8.78503 

0.0014749 

679 

461041 

313046839 

26.05763 

8.78935 

0.0014728 

680 

462400 

314432000 

26.07681 

8.79366 

0.0014706 

681 

463761 

315821241 

26.09598 

8.79797 

0.0014684 

682 

465124 

317214568 

26.11513 

8.80227 

0.0014663 

683 

466489 

318611987 

26.13427 

8.80657 

0.0014641 

684 

467856 

320013504 

26.15339 

8.81087 

0.0014620 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


ss 


n 

n2 

n3 

Vn 

vnr 

1 

n 

685 

469225 

321419125 

26.17250 

8.81516 

0.0014599 

686 

470596 

32282885(5 

26.19160 

8.81945 

0.0014577 

687 

471969 

324242703 

26.21068 

8.82373 

0.0014556 

688 

473344 

325660672 

26.22975 

8.82801 

0.0014535 

689 

474721 

327082769 

26.24881 

8.83229 

0.0014514 

690 

476100 

328509000 

26.26785 

8.83656 

0.0014493 

691  i 

477481 

329939371 

26.28688 

8.84082 

0.0014472 

692 

478864 

331373888 

26.30589 

8.84509 

0.0014451 

693 

480249 

332812557 

26.32489 

8.84934 

0.0014430 

694  i  481636 

334255384 

26.34388 

8.85360 

0.0014409 

695    483025 

335702375 

26.36285 

8.85785 

0.0014388 

696 

484416 

337153536 

26.38181 

8.86210 

0.0014368 

697  ; 

485809 

338608873 

26.40076 

8.86634 

0.0014347 

698 

487204 

340068392 

26.41969 

8.87058 

0.0014327 

699 

488601 

341532099 

26.43861 

8.87481 

0.0014306 

700 

490000 

343000000 

26.45751 

8.87904 

0.0014286 

701 

491401 

344472101 

26.47640 

8.88327 

0.0014265 

702 

492804 

345948408 

26.49528 

8.88749 

0.0014245 

703 

494209 

347428927 

26.51415 

8.89171 

0.0014225 

704 

495616 

348913664 

26.53300 

8.89592 

0.0014205 

705 

497025 

350402625 

26.55184 

8.90013 

0.0014184 

706 

498436 

351895816 

26.57066 

8.90434 

0.0014164 

707 

499849 

353393243 

26.58947 

8.90854 

0.0014144 

708 

501264 

354894912 

26.60817 

8.91274 

0.0014124 

709 

502681  i  356400829 

26.62705 

8.91693 

0.0014104 

710 

504100 

357911000 

26.64583 

8.92112 

0.0014085 

711 

505521 

359425431 

26.66458 

8.92531 

0.0014065 

712 

506944 

360944128 

26.68333 

8.92949 

0.0014045 

713 

508369 

362467097 

26.70206 

8.93367 

0.0014025 

714 

509796 

363994344 

26.72078 

8.93784 

0.0014006 

715 

511225 

365525875 

26.73948 

8.94201 

0.0013986 

716 

512656 

367061696 

26.75818 

8.94618 

0.0013966 

717 

514089 

368601813 

26.77686 

8.95034 

0.0013947 

718 

515524 

370146232 

26.79552 

8.95450 

0.0013928 

719 

516961 

371694959 

26.81418 

8.95866 

0.0013908 

720 

518400 

373248000 

26.83282 

8.96281 

0.0013889 

721 

519841 

374805361 

26.85144 

8.96696 

0.0013870 

722 

521284 

37(5367048 

26.8700(5 

8.97110 

0.0013850 

723 

522729 

377933067 

26.88866 

8.97524 

0.0013831 

724 

524176 

379503424 

26.90725 

8.97938 

0.0013812 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

W2 

n3 

V  n 

V-T 

1 

n 

725 

525625 

381078125 

26.92582 

8.98351 

0.0013793 

726 

527076 

382657176 

26.94439 

8.98764 

0.0013774 

727 

528529 

384240583 

26.96294 

8.99176 

0.0013755 

728 

529984 

385828352 

26.98148 

8.99589 

0.0013736 

729 

531441 

387420489 

27 

9 

0.0013717 

730 

532900 

389017000 

27.01851 

9.00411 

0.0013699 

731 

534361 

390617891 

27.03701 

9.00822 

0.0013680 

732 

535824 

392223168 

27.05550 

9.01233 

0.0013661 

733 

537289 

393832837 

27.07397 

9.01643 

0.0013643 

734 

538756 

395446904 

27.09243 

9.02053 

0.0013624 

735 

540225 

397065375 

27.11088 

9.02462 

0.0013605 

736 

541696 

398688256 

27.12932 

9.02871 

0.0013587 

737 

543169 

400315553 

27.14771 

9.03280 

0.0013569 

738 

544644 

401947272 

27.16616 

9.03689 

0.0013550 

739 

546121 

403583419 

27.18455 

9.04097 

0.0013532 

740 

547600 

405224000 

27.20291 

9.04504 

0.0013514 

741 

549081 

406869021 

27.22132 

9.04911 

0.0013495 

742 

550564 

408518488 

27.23968 

9.05318 

0.0013477 

743 

552049 

410172407 

27.25803 

9.05725 

0.0013459 

744 

553536 

411830784 

27.27636 

9.06131 

0.0013441 

745 

555025 

413493625 

27.29469 

9.06537 

0.0013423 

746 

556516 

415160936 

27.31300 

9.06942 

0.0013405 

747 

558009 

416832723 

27.33130 

9.07347 

0.0013387 

748 

559504 

418508992 

27.34959 

9.07752 

0.0013369 

749 

561001 

420189749 

27.36786 

9.08156 

0.0013351 

750 

562500 

421875000 

27.38613 

9.08560 

0.0013333 

751 

564001 

423564751 

27.40438 

9.08964 

0.0013316 

752 

565504 

425259008 

27.42262 

9.09367 

0.0013298 

753 

567009 

426957777 

27.44085 

9.09770 

0.0013280 

754 

568516 

428661064 

27.45906 

9.10173 

0.0013263 

755 

570025 

430368875 

27.47726 

9.10575 

0.0013245 

756 

571536 

432081216 

27.49545 

9.10977 

0.0013228 

757 

573049 

433798093 

27.51363 

9.11378 

0.0013210 

758 

574564 

435519512 

27.53180 

9.11779 

0.0013193 

759 

576081 

437245479 

27.54995 

9.12180 

0.0013175 

760 

577600 

438976000 

27.56810 

9.12581 

0.0013158 

761 

579121 

440711081 

27.58623 

9.12981 

0.0013141 

762 

580644 

442450728 

27.60435 

9.13380 

0.0013123 

763 

582169 

444194947 

27.62245 

9.13780 

0.0013106 

764 

583696 

445943744 

27.64055 

9.14179 

0.0013089 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


57 


n 

n2 

n3 

VT 

3 

Vn 

1 

n 

765 

585225 

447697125 

27.65863 

9.14577 

0.0013072 

766 

586756 

449455096 

27.67671 

9.14976 

0.0013055 

767 

588289 

451217663 

27.69476 

9.15374 

0.0013038 

768 

589824 

452984832 

27.71281 

9.15771 

0.0013021 

769 

591361 

454756609 

27.73085 

9.16169 

0.0013004 

770 

592900 

456533000 

27.74887 

9.16566 

0.0012987 

771 

594441 

458314011 

27.76689 

9.16962 

0.0012970 

772 

595984 

460099648 

27.78489 

9.17359 

0.0012953 

773 

597529 

461889917 

27.80288 

9.17754 

0.0012937 

774 

599076 

463684S24 

27.82086 

9.18150 

0.0012920 

775 

600625 

465484375 

27.83882 

9.18545 

0.0012903 

776 

602176 

467288576 

27.85678 

9.18940 

0.0012887 

777 

603729 

469097433 

27.87472 

9.19335 

0.0012870 

778 

605284 

470910952 

27.89265 

9.19729 

0.0012853 

779 

606841 

472729139 

27.91057 

9.20123 

0.0012837 

780 

608400 

474552000 

27.92848 

9.20516 

0.0012821 

781 

609961 

476379541 

27.94638 

9.20910 

0.0012804 

782 

611524 

478211768 

27.96426 

9.21303 

0.0012788 

783 

613089 

480048687 

27.98214 

9.21695 

0.0012771 

784 

614656 

481890304 

28 

9.22087 

0.0012755 

785 

616225 

483736625 

28.01785 

9.22479 

0.0012739 

786 

617796 

485587656 

28.03569 

9.22871 

0.0012723 

787 

619369 

487443403 

28.05352 

9.23262 

0.0012706 

788 

620944 

489303872 

28.07134 

9.22653 

0.0012690 

789 

622521 

491169069 

28.08914 

9.24043 

0.0012674 

790 

624100 

493039000 

28.10694 

9.24434 

0.0012658 

791 

625681 

494913671 

28.12472 

9.24823 

0.0012642 

792 

627264 

496793088 

28.14249 

9.25213 

0.0012629 

793 

628849 

498677257 

28.16026 

9.25602 

0.0012610 

794 

630436 

500566184 

28.17801 

9.25991 

0.0012594 

795 

632025 

502459875 

28.19574 

9.26380 

0.0012579 

796 

633616 

504358336 

28.21347 

9.26768 

0.0012563 

797 

635209 

506261573 

28.23119 

9.27156 

0.0012547 

798 

!  636804 

508169592 

28.24889 

9.27544 

0.0012531 

799 

638401 

510082399 

28.26659 

9.27931 

0.0012516 

800 

640000 

512000000 

28.28427 

9.28318 

0.0012500 

801 

641601 

513922401 

28.30194 

9.28704 

0.0012484 

802 

643204 

515849608 

28.31960 

9.29091 

0.0012469 

803 

644809 

517781627 

28.33725 

9.29477 

0.0012453 

804 

646416 

519718464 

28.35489 

9.29862 

0.0012438 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

n2 

W3 

V  n 

V^T 

1 

n 

805 

648025 

521660125 

28.37252 

9.30248 

0.0012422 

806 

649636 

523606616 

28.39014 

9.30638 

0.0012407 

807 

651249 

525557943 

28.40775 

9.31018 

0.0012392 

808 

652864 

527514112 

28.42534 

9.31402 

0.0012376 

809 

654481 

529475129 

28.44293 

9.31786 

0.0012361 

810 

656100 

531441000 

28.46050 

9.32170 

0.0012346 

811 

657721 

533411731 

28.47806 

9.32553 

0.0012330 

812 

659344 

535387328 

28.49561 

9.32936 

0.0012315 

813 

660969 

537367797 

28.51315 

9.33319 

0.0012300 

814 

662596 

539353144 

28.53069 

9.33702 

0.0012285 

815 

664225 

541343375 

28.54820 

9.34084 

0.0012270 

816 

665856 

543338496 

28.56571 

9.34466 

0.0012255 

817 

667489 

545338513 

28.58321 

9.34847 

0.0012240 

818 

669124 

547343432 

28.60070 

9.35229 

0.0012225 

819 

670761 

549353259 

28.61818 

9.35610 

0.0012210 

820 

672400 

551368000 

28.63564 

9.35990 

0.0012195 

821 

674041 

553387661 

28.65310 

9.36270 

0.0012180 

822 

675684 

555412248 

28.67054 

9.36751 

0.0012165 

823 

677329 

557441767 

28.68798 

9.37130 

0.0012151 

824 

678976 

559476224 

28.70540 

9.37510 

0.0012136 

825 

680625 

56  J  51  5625 

28.72281 

9.37889 

0.0012121 

826 

682276 

563559976 

28.74022 

9.38268 

0.0012107 

827 

683929 

565609283 

28.75761 

9.38646 

0.0012092 

828 

685584 

567663552 

28.77499 

9.39024 

0.0012077 

829 

687241 

569722789 

28.79236 

9.39402 

0.0012063 

830 

688900 

571787000 

28.80972 

9.39780 

0.0012048 

831 

690561 

573856191 

28.82707 

9.40157 

0.0012034 

832 

692224 

575930368 

28.84441 

9.40534 

0.0012019 

833 

693889 

578009537 

28.86174 

9.40911 

0.0012005 

834 

695556 

580093704 

28.87906 

9.41287 

0.0011990 

835 

697225 

582182875 

28.89637 

9.41663 

0.0011976 

836 

698896 

584277056 

28.91366 

9.42039 

0.0011962 

837 

700569 

586376253 

28.93095 

9.42414 

0.0011947 

838 

702244 

588480472 

28.94823 

9.42789 

0.0011933 

839 

703921 

590589719 

28.96550 

9.43164 

0.0011919 

840 

705600 

592704000 

28.98275 

9.43538 

0.0011905 

841 

707281 

594823321 

29 

9.43913 

0.0011891 

842 

708964 

596947688 

29.01724 

9.44287 

0.0011876 

843 

710649 

599077107   29.03446 

9.44661 

0.0011862 

844 

712336 

601211584  I  29.05168    9.45034 

0.0011848 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


59 


n 

n2 

n3 

Vn 

^ 

1 

n 

845 

714025 

603351125 

29.06888 

9.45407 

0.0011834 

846 

715710 

605495736 

29.08608 

9.45780 

0.0011820 

847 

717409 

(507645423 

29.10326 

9.46152 

0.0011806 

848 

719104 

609800192 

29.12044 

9.46525 

0.0011792 

849 

720801 

611900049 

29.13760 

9.40897 

0.0011779 

850 

722500 

014125000 

29.1547(5 

9.47268 

0.0011765 

851 

724201 

010295051 

29.17190 

9.47640 

0.0011751 

852 

725904 

618470208 

29.18904 

9.48011 

0.0011737 

853 

727009 

020(550477 

29.20610 

9.48381 

0.0011723 

854 

729310 

622835804 

29.22328 

9.48752 

0.0011710 

855 

731025 

025020375 

29.24038 

9.49122 

0.0011696 

850 

73273(5 

027222010 

29.25748 

9.49492 

0.0011682 

857 

734449 

029422793 

29.27456 

9.49861 

0.0011669 

858 

730104 

031028712 

29.29164 

9.50231 

0.0011655 

859 

737881 

033839779 

29.30870 

9.50600 

0.0011641 

800 

739000 

030050000 

29.32576 

9.50969 

0.0011628 

801 

741321 

03X277381 

29.34280 

9.51337 

0.0011614 

802 

743044 

64050392S 

29.35984 

9.51705 

0.0011601 

868 

744709 

042735047 

29.37686 

9.52073 

0.001  1587 

H54 

740490 

044972544 

29.39388 

9.52441 

0.0011574 

805 

748225 

647214025 

29.41088 

9.52808 

0.0011501 

800 

749950 

049401890 

29.42788 

9.53175 

0.0011547 

807 

751089 

051714303 

29.44486 

9.53542 

0.0011534 

8(58 

753424 

053972032 

29.40184 

9.53908 

0.0011521 

809 

755101 

050234909 

29.47881 

9.54274 

0.0011507 

870 

750900 

058503000 

29.49570 

9.54640 

0.0011494 

871 

758041 

000770311 

29.51271 

9.55006 

0.0011481 

872 

700384 

663054848 

29.52905 

9.55371 

0.0011408 

873 

702129 

665338617 

29.54057 

9.55736 

0.0011455 

874 

703870 

667627624 

^9.56349 

9.56101 

0.0011442 

875 

765025 

669921875 

29.58040 

9.56466 

0.0011429 

870 

767376 

672221376 

29.59730 

9.56830 

0.0011410 

877 

769129 

674526133 

29.61419 

9.57194 

0.0011403 

878 

770884 

676830152 

29.63106 

9.57557 

0.0011390 

879 

772641 

679151439 

29.64793 

9.57921 

0.0011377 

880 

774400 

681472000 

29.66479 

9.58284 

0.0011364 

881 

776161 

OS3797841 

29.68164 

9.58647 

0.0011351 

882 

777924  686128968 

29.69848 

9.59009 

0.0011338 

883 

779688 

088405387 

29.71532 

9.59372 

0.0011325 

884 

781450 

690807104 

29.73214 

9.59734 

0.0011312 

6o 


SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

n2 

n3 

V  n 

v^T 

1 

n 

885 

783225 

693154125 

29.74895 

9.60095 

0.0011299 

886 

784996 

695506456 

29.76575 

9.60457 

0.0011287 

887 

786769 

697864103 

29.78255 

9.60818 

0.0011274 

888 

788544 

700227072 

29.79933 

9.61179 

0.0011261 

889 

790321 

702595369 

29.81610 

9.61540 

0.0011249 

890 

792100' 

704969000 

29.83287 

9.619 

0.0011236 

891 

793881 

707347971 

29.84962 

9.62260 

0.0011223 

892 

795664 

709732288 

29.86637 

9.62620 

0.0011211 

893 

797449 

712121957 

29.88311 

9.62980 

0.0011198 

894 

799236 

714516984 

29.89983 

9.63339 

0.0011186 

895 

801025 

716917375 

29.91655 

9.63698 

0.0011173 

896 

802816 

719323136 

29.93326 

9.64057 

0.0011161 

897 

804609 

721734273 

29.94996 

9.64415 

0.0011148 

898 

806404 

724150792 

29.96665 

9.64774 

0.0011136 

899 

808201 

726572699 

29.98333 

9.65132 

0.0011123 

900 

810000 

729000000 

30 

9.65489 

0.0011111 

901 

811801 

731432701 

30.01666 

9.65847 

0.0011099 

902 

813604 

733870808 

30.03331 

9.66204 

0.0011086 

903 

815409 

736314327 

30.04996 

9.66561 

0.0011074 

904 

817216 

738763264 

30.06659 

9.66918 

0.0011062 

905 

819025 

741217625 

30.08322 

9.67274 

0.0011050 

906 

820836 

743677416 

30.09983 

9.67630 

0.0011038 

907 

822649 

746142643 

30.11644 

9.67986 

0.0011025 

908 

824464 

748613312 

30.13304 

9.68342 

0.0011013 

909 

826281 

751089429 

30.14963 

9.68697 

0.0011001 

910 

828100 

753571000 

30.16621 

9.69052 

0.0010989 

911 

829921 

756058031 

30.18278 

9.69407 

0.0010977 

912 

831744 

758550528 

30.19934 

9.69762 

0.0010965 

913 

833569 

761048497 

30.21589 

9.70116 

0.0010953 

914 

835396 

763551944 

30.23243 

9.70470 

0.0010941 

915 

837225 

766060875 

30.24897 

9.70824 

0.0010929 

916 

839056 

768575296 

30.26549 

9.71177 

0.0010917 

917 

840889 

771095213 

30.28201 

9.71531 

0.0010905 

918 

842714 

773620632 

30.29851 

9.71884 

0.0010893 

919 

844561 

776151559 

30.31501 

9.72236 

0.0010881 

920 

846400 

778688000 

30.33150 

9.72589 

0.0010870 

921 

848241 

781229961 

30.34798 

9.72941 

0.0010858 

922 

850084 

783777448 

30.36445 

9.73293 

0.0010846 

923 

851929 

786330467 

30.38092 

9.73645 

0.0010834 

924 

853776 

788889024 

30.39737 

9.73996 

0.0010823 

SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS.  6 1 


n 

n2 

n3 

\fn~ 

<hr 

1 

n 

925 

855625 

791453125 

30.41381 

9.74348 

0.0010811 

926 

857476 

794022776 

30.43025 

9.74699 

0.0010799 

927 

859329 

796597983 

30.44667 

9.75049 

0.0010787 

928 

861184 

799178752 

30.46309 

9.75400 

0.0010776 

929 

863041 

801765089 

30.47950 

9.75750 

0.0010764 

930 

864900 

804357000 

30.49590 

9.76100 

0.0010753 

931 

866761 

806954491 

30.51229 

9.76450 

0.0010741 

932 

868624 

809557568 

30.52868 

9.76799 

0.0010730 

933 

870489 

812166237 

30.54505 

9.77148 

0.0010718 

934 

872356 

814780504 

30.56141 

9.77497 

0.0010707 

935 

874225 

817400375 

30.57777 

9.77846 

0.0010695 

936 

876096 

820025856 

30.59412 

9.78295 

0.0010684 

937 

877969 

822656953 

30.61046 

9.78543 

0.0010672 

938 

879844 

825293672 

30.62679 

9.78891 

0.0010661 

939 

881721 

827936019 

30.64311 

9.79239 

0.0010650 

940 

883600 

830584000 

30.65942 

9.79586 

0  0010638 

941 

885481 

833237621 

30.67572 

9.79933 

0.0010627 

942 

887364 

835896888 

30.69202 

9.80280 

0.0010616 

943 

889249 

838561807 

30.70831 

9.80627 

0.0010604 

944 

891136 

841232384 

30.72458 

9.80974 

0.0010593 

945 

893025 

843908625 

30.74085 

9.81320 

0.0010582 

946 

894916 

846590536 

30.75711 

9.81666 

0.0010571 

947 

896809 

849278123 

30.77337 

9.82012 

0.0010560 

948 

898704 

851971392 

30.78961 

9.82357 

0.0010549 

949 

900601 

854670349 

30.80584 

9.82703 

0.0010537 

950 

902500 

857375000 

30.82207 

9.83048 

0.0010526 

951 

904401 

860085351 

30.83829 

9.83392 

0.0010515 

952 

906304 

862801408 

30.85450 

9.83737 

0.0010504 

953 

908209 

865523177 

30.87070 

9.84081 

0.0010493 

954 

910116 

868250664 

30.88689 

9.84425 

0.0010482 

955 

912025 

870983875 

30.90307 

9.84769 

0.0010471 

956 

913936 

873722816 

30.91925 

9.85113 

0.0010460 

957 

915849 

876467493 

30.93542 

9.85456 

0.0010449 

958 

917764 

879217912 

30.95158 

9.85799 

0.0010438 

959 

919681 

881974079 

30.96773 

9.86142 

0.0010428 

960 

921600 

887436000 

30.98387 

9.86485 

0.0010417 

961 

923521 

887503681 

31 

9.86827 

0.0010406 

962 

925444 

890277128 

31.01612 

9.87169 

0.0010395 

963 

927369 

893056347 

31.03224 

9.87511 

0.0010384 

964 

929296 

895841344 

31.04835 

9.87853 

0.0010373 

62 


SQUARES,    CUBES,    ROOTS,    AND    RECIPROCALS. 


n 

n2 

n3 

+/~n 

£v 

1 

n 

965 

931225 

898632125 

31.06445 

9.88195 

0.0010363 

966 

933156 

901428696 

31.08054 

9.88536 

0.0010352 

967 

935089 

904231063 

31.09662 

9.88877 

0.0010341 

968 

937024 

907039232 

31.11270 

9.89217 

0.0010331 

969 

938961 

909853209 

31.12876 

9.89558 

0.0010320 

970 

940900 

912673000 

31.14482 

9.89898 

0.0010309 

971 

942841 

915498611 

31.16087 

9.90238 

0.0010299 

972 

944788 

918330048 

31.17691 

9.90578 

0.0010288 

973 

946729 

921167317 

31.19295 

9.90918 

0.0010277 

974 

948676 

924010424 

31.20897 

9.91257 

0.0010267 

975 

950625 

926859375 

31.22499 

9.91596 

0.0010256 

976 

952576 

929714176 

31.24100 

9.91935 

0.0010246 

977 

954529 

932574833 

31.25700 

9.92274 

0.0010235 

978 

956484 

935441352 

31.27299 

9.92612 

0.0010225 

979 

958441 

938313739 

31.28898 

9.92950 

0.0010215 

980 

960400 

941192000 

31.30495 

9.93288 

0.0010204 

981 

|  962361 

944076141 

31.32092 

9.93626 

0.0010194 

982 

964324 

946966168 

31.33688 

9.93964 

0.0010183 

983 

966289 

949862087 

31.35283 

9.94301 

0.0010173 

984 

968256 

952763904 

31.36877 

9.94638 

0.0010163 

985 

970225 

955671625 

31.38471 

9.94975 

0.0010152 

986 

972196 

958585256 

31.40064 

9.95311 

0.0010142 

987 

974169 

961504803 

31.41656 

9.95648 

0.0010132 

988 

976144 

964430272 

31.43247 

9.95984 

0.0010121 

989 

978121 

967361669 

31.44837 

9.96320 

0.0010111 

990 

980100 

970299000 

31.46427 

9.96655 

0.0010101 

991 

982081 

973242271 

31.48015 

9.96991 

0.0010091 

992 

984064 

976191488 

31.49603 

9.97326 

0.0010081 

993 

986049 

979146657 

31.51190 

9.97661 

0.0010070 

994 

988036 

982107784 

31.52777 

9.97996 

0.0010060 

995 

990025 

985074875 

31.54362 

9.98331 

0.0010050 

996 

992016 

988047936 

31.55947 

9.98665 

0.0010040 

997 

994009 

991026973 

31.57531 

9.98999 

0.0010030 

998 

996004 

994011992 

31.59114 

9.99333 

0.0010020 

999 

998001 

997002999 

31.60696 

9.99667 

0.0010010 

IRotes  on  Hlgebra. 


Algebra  is  that  branch  of  mathematics  in  which  the  quan- 
tities are  denoted  by  letters  and  the  operations  to  be  performed 
upon  them  are  indicated  by  signs.  The  same  signs  are  used  to 
indicate  the  same  operations  as  in  arithmetic. 

Signs  of  Quantity  and  Signs  of  Operation. 

If  a  quantity  is  written  a  -f  (  —  £),  the  sign  that  precedes  the 
parenthesis  is  called  the  sign  of  operation,  and  the  sign  within 
the  parenthesis  is  called  the  sign  of  quantity  with  respect  to  b, 
but  expressions  of  this  kind  can  be  reduced  to  have  only  one 
sign.  Thus,  a  -f  (  —  ft)  =  a  —  b  and  this  final  sign  is  called  the 
essential  sign. 

a  +  (+  b}  =  a  +  b. 

a  --  (  —  b)  =  a  —  b. 


Thus,  when  the  sign  of  operation  and  the  sign  of  quantity 
are  alike  the  essential  sign  is  +,  but  if  they  are  unlike  the  es- 
sential sign  is  —  . 

In  multiplying  any  two  quantities,  like  signs  in  the  two 
factors  give  4-  in  the  product,  but  unlike  signs  in  the  two  factors 
give  —  in  the  product  ;  thus,  (-f  a)  X  (  —  b}  =  —  ab,  and  (  —  a] 
X  (—  b)  =  ab. 

In  division,  like  signs  in  dividend  and  divisor  give  -f-  in  the 
quotient,  but  unlike  signs  in  dividend  and  divisor  give  —  in  the 
quotient;  thus: 

—  a  a 


Useful  Formulas  and  Rules  in  Algebra. 

The  following  rules  are  very  useful  to  remember  in  solv- 
ing practical  problems  in  algebra.  Let  a  and  b  represent  any 
two  quantities ;  then  a  +  b  will  represent  their  sum  and  a  —  b 
their  difference;  then  (a  +  b)  X  (a  -f  b)  =  a2  +  2  ab  +  £2. 

(a  +  b)  X  (a  +  b)  is  also  written  (a  +  £)«. 
(63) 


64  NOTES    ON   ALGEBRA. 

This  rule  reads: 

The  square  of  the  sum  of  any  two  quantities  is  equal  to 
the  square  of  the  first  quantity  plus  double  the  product  of  both 
quantities,  plus  the  square  of  the  second  quantity. 


This  rule  reads: 

The  square  of  the  difference  of  any  two  quantities  is  equal 
to  the  square  of  the  first  quantity  minus  twice  the  product  of 
both  quantities,  plus  the  square  of  the  second  quantity. 

(a  +  b)  X  («  —  b]  =  a2  —  P. 
This  rule  reads : 

The  sum  of  any  two  quantities  multiplied  by  their  differ- 
ence is  equal  to  the  difference  of  their  squares. 

Extracting  Roots. 

An  even  root  of  a  positive  quantity  is  either  -f  or  — .  An 
even  root  cannot  be  extracted  of  a  negative  quantity,  as  \/~a* 
may  be  either  a  or  —  a ;  but  -v —  a1  is  impossible,  because 
(—  a)  X  (—  a)  =  a'2  and  (+  a)  X  (+  a)  =  a2. 

An  odd  root  may  be  extracted  as  well  of  a  negative  quantity 
as  a  positive  quantity,  and  the  sign  of  the  root  is  always  the 
same  as  the  sign  of  the  quantity  before  the  root  was  extracted. 

3  3  

Thus :    V  a3  =  a,  but  V  (—  a)3  =  (— a). 

Powers. 

When  a  number  or  a  quantity  is  to  be  multiplied  by  itself 
a  given  number  of  times,  the  operation  is  indicated  by  a  small 
number  at  the  right-hand  corner  of  the  quantity ;  for  instance, 
0-2  =  0  x  a. 

A  quantity  of  this  kind  is  called  a  power;  the  small  number 
is  called  the  exponent,  or  the  index  of  the  power.  Two  powers 
of  the  same  kind  may  be  multiplied  by  adding  the  exponents ; 
for  instance,  a2  a*  =  a'2+ 3  =  ab. 

Two  powers  of  the  same  kind  may  be  divided  by  subtract- 
ing their  exponents ;  for  instance, 

J^.    =    05-2  =  03  =  a   X   0  X   a. 

a2 
ab 

-^   =    05-3    =  02  =    a   X   a. 

4  =  a1  =  a. 


NOTES    ON   ALGEBRA.  65 

Thus,  any  quantity  in  0  power  must  be  1,  because  always 
when  dividend  and  divisor  are  alike  the  quotient  must  be  1. 

—  =  fl5-6  =  a~l,  but  a5  divided  by  a5  is  equal  to  1 ;    therefore 

4  must  be  -L  ;      4  =  <r*  _  J_. 
a6  a  a1  a2 

Thus,  any  quantity  with  a  negative  exponent  is  one  divided 
by  that  quantity  considering  the  exponent  as  positive.  We 
may,  therefore,  say  that  as  a  positive  exponent  indicates  how 
many  times  a  quantity  is  to  be  used  as  a  factor,  a  negative  ex- 
ponent indicates  how  many  times  a  quantity  should  be  used  as 
a  divisor ;  for  instance, 

<r*  = 


a   '  a  X  a    '  a  Xa  X  a 

Thus : 

6-1=1;          o-a  =  (1)2  =  ^  .          3-2  =  (1)2  =  ^  etC. 

Equations. 

An  algebraic  expression  of  equality  between  two  quantities 
is  called  an  equation.  The  two  quantities  are  connected  by  the 
sign  of  equality,  and  the  quantity  on  the  left-hand  side  of  the 
sign  is  called  the  first  member,  and  the  quantity  on  the  right- 
hand  side  is  called  the  second  member  of  the  equation. 

If  a  part  or  all  of  the  known  quantities  are  expressed  in 
letters  it  is  called  a  literal  equation:  ax  =  a  +  b  —  2c  is  a 
literal  equation. 

If  the  equation  contains  no  letters  except  the  unknown 
quantity,  usually  expressed  by  x,  it  is  called  a  numerical  equa- 
tion ;  for  instance,  5  x  =  12  —  7  +  9,  is  a  numerical  equation. 

In  solving  equations,  we  may,  without  destroying  the  equal- 
ity of  the  equation,  add  an  equal  quantity  to  both  members, 
subtract  an  equal  quantity  from  both  members,  multiply  both 
members  by  an  equal  quantity,  divide  both  members  by  an 
equal  quantity,  extract  the  same  root  of  both  members,  raise 
both  members  to  the  same  power. 

Quantities  inclosed  by  parentheses,  a  bar,  or  under  a  radical 
sign,  and  quantities  connected  by  the  sign  of  multiplication, 
must  always  be  considered  and  operated  upon  as  one  quantity. 

EXAMPLE  1. 

*  =  3XS  +  10  —  3  +  3X10. 
x  =  24  +  10  —  3  +  30. 
x  =  64— 3. 
x  =  61. 


66  NOTES  ON  ALGEBRA. 

EXAMPLE  2. 

3  +  8  +  10 
*=      ~W~ 
21 


EXAMPLE  3. 

X 


x  =12  X 

x  =  12  X  (    7— 
x  =92 
EXAMPLE  4. 


.Wx(^if2j 

fVo) 


8  —  3        8  +  10 
.*  -  =  12  X       3  3 

,  JL     jt§_ 

^*"  ==  12  X     o    +    o 

»J  O 

or  =  20  +  6 

x  =26 


EXAMPLE  5. 

X 


=  (  5±?  x   (8  —  3)  +  \/20  +  ie)  X  2 
=  (— |-  X  5  +  \/36")  X  2 


r=  (6X5  +  6)  X  2 

r  ==  36  X  2 


When  an  equation  consists  of  more  than  one  unknown  quan- 
tity, as  many  equations  may  be  arranged  as  there  are  unknown 
quantities,  and  one  equation  is  solved  so  that  the  value  of  one  of 
its  unknown  quantities  is  expressed  in  terms  of  the  other,  and 
this  value  is  substituted  in  the  other  equation. 

EXAMPLE. 

Two  shafts  are  to  be  connected  by  two  gears  of  16  diam- 
etral pitch  ;  the  distance  between  centers  is  6^  inches.  The 
ratio  of  gearing  shall  be  1  to  3.  How  many  teeth  in  each  gear  ? 


NOTES  ON   ALGEBRA.  67 

Call  small  gear  x  and  large  gear  y  ;  then  x  +  y  must  be  108, 
because  6%  X  16  =  108,  which  is  the  number  of  teeth  in  both 
gears  added  together.      The  ratio  is  1  to  3;   therefore  §x  =  y. 
Thus: 

x  +  y  =  108,  transposed  to 
x  +  3  x  =  108. 
4  x  =  108. 


jr  =  27  teeth  for  small  gear. 
The  large  gear  =  27  X  3  =  81  teeth. 

Quadratic  Equations. 

Equations  containing  one  or  more  unknown  quantities  in 
the  second  power  are  called  quadratic  equations.  If  the  un- 
known quantity  only  exists  in  the  second  power  the  equation 
may  be  brought  to  the  form  x2  =  a  and  x  =  /v/~0T 

This  square  root  may  be  either  plus  or  minus. 

If  the  unknown  quantity  exists  in  both  the  first  and  the 
second  power  the  equation  may  be  brought  to  the  form  x*  -{-a  x 
=  b,  or  it  may  be  brought  to  the  form  x2  —  a  x  =  b. 

The  coefficient  a  may  be  any  number.  After  the  equation 
is  brought  to  this  form,  complete  the  square  of  the  first  member 
by  adding  to  both  members  of  the  equation  the  square  of  half 
the  coefficients;  this  will  make  the  left  member  of  the  equa- 
tion a  complete  square. 

EXAMPLE. 

A  coal  bin  is  to  hold  six  tons  of  coal.  Allow  40  cubic 
feet  per  ton.  (It  takes  35  to  40  cubic  feet  to  hold  a  ton  of 
coal  in  a  bin).  Make  the  width  of  the  bin  6  feet,  and  the 
length  equal  to  the  width  and  the  depth  added  together.  How 
deep  and  how  long  will  the  bin  be  ? 

Depth  =  x  and  length  =  y. 

6  x  y  =  240,  because  6  times  40  equals  240. 

6  +  x  =  y,  because  width  +  depth  =  length. 

Thus: 

6  x  (6  +  *)  =  240. 
6  **  +  36  x  =  240. 

Dividing  by  6  we  have  :  x*  +  Qx  =  40. 

Completing  the  square  :  x*  +  Qx  +  32  =  40  +  3*. 

x  +  3  =  V  40  +  9. 

Extracting  the  square  root:  x  +  3  =  \/  49. 

*•  +  3  =  7. 
x  =  7  —  3. 
x  =  4  feet  deep. 


68  NOTES    ON   ALGEBRA. 

Length  =  width  +  depth,  =6+4,  =10  feet. 
This  satisfies  the  conditions  of  the  problem,  because 

6  X  4  X  10  =  240  cubic  feet, 
and  the  width  and  depth  added  together  equal  the  length. 

Progressions. 

A  progression  is  a  series  of  numbers  increasing  or  decreas- 
ing, according  to  a  fixed  law. 

The  successive  numbers  of  which  the  progression  consists 
are  called  terms  ;  the  first  and  the  last  terms  are  called  the 
extremes  and  the  others  are  called  the  means. 

ARITHMETICAL  PROGRESSION. 

An  arithmetical  progression  is  a  series  of  numbers  which 
increase  or  decrease  by  a  constant  difference.  For  instance  : 

2,  4,  6,  8,  10,  12,  14,  16,  is  an  ascending  series  ; 
20,  18,  16,  14,  12,  10,  8,  is  a  descending  series. 
In  each  of  these  series  the  common  difference  is  2. 

The  following  are  the  elements  considered  in  an  arith- 
metical progression  : 

a  —  First  term  ;  /  =  Ic.st  term  ;  d  =  the  common  difference  ; 
n  •=.  the  number  of  terms  ;  j  =  the  sum  of  all  the  terms. 

When  any  three  of  these  quantities  are  known  the  other  two 
may  be  calculated  : 

In  the  above  example  of  an  ascending  series  : 
a  —  2  ;     /  =  16  ;     d—  2  ;     ;/  =  8  ;     s  =  72. 

Formulas  :  Examples  : 

a  —  I—  (;/  —  !)  X  d  a  —  16  —  (8  —  1)  X  2  =  2 

l=a  +  (n  —  1)  X  d  /=2  +  (S  —  1)  X  2  =  16 


—  1 


_  (2  +  16)  X  8   _ 


NOTES   ON   ALGEBRA.  69 

In  the  above  example  of  a  descending  series  : 

a  =  20  ;    /  =  8  ;    d  —  2  ;    »  =  7  ;    J  =  98. 
Formulas  :  Examples  : 

a  —  /  -f  (  «  —  i)x^/  «  =  8  +  (7  —  1)  X  2  =  20 

l=a  —  (n  —  l)Xd  /=20  —  (7  —  1)  X  2  =  8 


—  1  7—1 


-7  X»  ,  =  2-°-+8X  7  =  98 


GEOMETRICAL   PROGRESSION. 

A  Geometrical  Progression  is  a  series  of  numbers  which 
increase  or  decrease  by  a  common  constant  ratio.  For 
instance : 

3,  6,  12,  24,  48,  is  an  ascending  series;  48,  24,  12,  6,  3  is  a 
descending  series. 

The  following  are  the  elements  considered  in  a  geometrical 
progression : 

a  =  first  term ;  /  =  last  term ;  r  =  ratio ;  n  =  number 
of  term ;  s  —  sum  of  the  terms. 

When  three  of  these  are  known  the  other  two  may  be  cal- 
culated. 

In  this  example  of  an  ascending  series  : 

a  —  3 ;    /  =  48  ;    r  =  2  ;    n  =  5 ;    s  =  93. 
Formulas :  Examples : 

a  =  _4§_  =  _48_  _  48_  _  3 
2  6-1         24    "      16 

l=aX  r*~l  /=3X2  *~l  =  3  X  16  =  48 

n  -  1  5-14 

„ 

log.  48  — 


_   log.  I— log,  a     , 


f_ 
r— 1  2—1 


log.  2 

1.681241  — 0.477121 +1_^ 
0.30103 

48  X  2  —  3  _ 


70  NOTES   ON  ALGEBRA. 

In  this  example  of  a  descending  series : 

a  ==  48 ;  /  =  3 ;    r  =  2 ;    n  =  5 ;    s  =  93. 

Formulas :  Examples  : 

asr/XrV1  a  =  3  X  25-!  =  3  X  16  =  48 

l=_a_  f=    48  _^   48   ^    48   _  3 

rn-i  2  6~1         24    "      16 

n-1    _  5-1                       4 


w=  fly.  48  —  log.  3 
log,  a  -log.  I  log.* 

lom  r  1.681241  —  0.477121  ,  l  _  5 


0.30103 


_aXr—l  48  X  2 

--         — 


The  Arithmetical  Mean. 

The  arithmetical  mean  of  two  or  more  quantities  is  obtained 
by  adding  the  quantities  and  dividing  the  sum  by  their  number. 
For  instance,  the  arithmetical  mean  of  14  and  16  is  14  +  16  —  ^5 

Thus  :  The  arithmetical  mean  is  simply  the  average. 

The  Geometrical  Mean. 

The  geometrical  mean  of  two  quantities  is  the  square  root 
of  their  product.  For  instance,  the  geometrical  mean  of  14  and 
16  is  \/14  X  16  =  14.9666. 

The  geometrical  mean  of  two  numbers  is  also  called  their 
mean  proportional. 

When  the  difference  between  two  numbers  is  small  as  com- 
pared to  either  of  them,  their  arithmetical  mean  is  approxi- 
mately equal  to  their  geometrical  mean. 

This  fact  may  be  used  to  advantage  for  calculating  approxi- 
mately a  root  of  any  number. 

For  instance,  find  the  square  root  of  148. 

Knowing  that  the  square  of  12  is  144,  twelve  is  used  as  a 
divisor,  thus  : 

148  A  12.333  +  12 

—  —  —  12.333,  and  -  -  -  =  12.166, 

which  is  correct  within  0.005. 


^Logarithms. 


Logarithms  are  a  series  of  numbers  computed  in  order  to 
facilitate  all  kinds  of  laborious  calculations,  such  as  evolution, 
involution,  multiplication  and  division. 

Addition  takes  the  place  of  multiplication,  subtraction  the 
place  of  division;  multiplication  that  of  involution,  and  divi- 
sion of  evolution. 

The  logarithm  of  any  given  number  is  the  exponent  of  the 
power  to  which  another  fixed  number,  called  the  base,  must  be 
raised  in  order  to  produce  the  given  number. 

There  are  two  systems  of  logarithms  in  more  or  less  general 
use  in  mechanical  calculations:  namely,  the  Napierian  system 
and  the  Briggs  system. 

The  Napierian  system  of  logarithms  was  invented  and 
tables  published  by  Baron  John  Napier,  a  Scotch  mathema- 
tician, in  1614,  but  these  tables  were  improved  by  John  Speidell 
in  1019. 

The  modulus  of  any  system  of  logarithms  is  a  constant  by 
which  the  Napierian  logarithm  of  any  given  number  must  be 
multiplied  in  order  to  obtain  the  logarithm  for  the  same  number 
in  the  other  system. 

The  base  of  the  Napierian  system  of  logarithms  is  an  in- 
commensurable number  expressed  approximately  by  2.718281828. 
In  mathematical  works  this  base  is  usually  denoted  by  the 
letter  e. 

The  Napierian  logarithms  are  frequently  called  hyperbolic 
logarithms,  from  their  relation  to  certain  areas  included  Between 
the  equilateral  hyperbola  and  its  asymptotes. 

The  Napierian  logarithms  are  sometimes  called  natural 
logarithms. 

The  Briggs  system  of  logarithms  was  first  invented  and 
computed  by  Professor  Henry  Briggs  of  London  in  1615,  and 
is  usually  termed  the  common  system  of  logarithms.  When- 
ever logarithms  in  general  is  mentioned  the  Briggs  system  is 
always  the  one  referred  to. 

The  Briggs  system  of  logarithms  has  for  its  modulus 
0.4342945,  and  10  for  its  base.  Therefore  the  Briggs  logarithm 
of  a  number  is  the  exponent  of  the  power  to  which  10  must  be 
raised  in  order  to  give  the  number.  Thus : 


LOGARITHMS. 


Log.  1=0  because  10°  =          1. 

"  10  =  1        "        101  =          10. 

"  100  =  2  "  102  =  100. 
«  1,000  =  3  "  103=  1,000. 
"  10,000  =  4  "  104  =  10,000. 

The  logarithm  of  any  number  between  1  and  10  is  a  frac- 
tion smaller  than  1.  The  logarithm  of  any  number  between  10 
and  100  is  a  number  between  1  and  2.  The  logarithm  of  any 
number  between  100  and  1000  is  a  number  between  2  and  3,  etc. 

The  decimal  part  of  the  logarithms  is  called  the  mantissa, 
and  is  given  in  the  table  commencing  on  page  88. 

The  integer  part  of  a  logarithm  is  called  the  index  or  some- 
times the  characteristic,  and  is  not  given  in  the  table,  but 
is  obtained  by  the  rule  that  it  is  one  less  than  the  number 
of  figures  in  the  integer  part  of  the  number ;  thus,  the  index  of  a 
logarithm  for  any  number  consisting  of  two  figures  must  be  1; 
the  index  of  the  logarithm  for  a  number  consisting  of  three 
figures  must  be  2,  etc. 

The  index  of  the  logarithm  of  a  decimal  fraction  is  a  nega- 
tive number.  Sometimes  the  negative  index  is  denoted  by 
writing  a  minus  sign  over  it;  for  instance,  log.  0.5240  =  1.719331, 
or  the  negative  index  is  denoted  by  writing  it  after  the  mantissa  ; 
thus,  log.  0.5240  =  9.719331  — 10.  This,  of  course,  is  of  exactly 
the  same  value  whether  written  —  1  or  9  —  10.  Either  of 
these  expressions  is  minus  one  in  value,  but  it  is  more  con- 
venient in  logarithmic  calculations  to  write  the  negative  index 

after  the  mantissa ;   thus,  instead  of  writing  1,   write  9 

—  10  ;  instead  of  2,  write  8 —  10,  etc.     Only  the  mantissa 

is  given  in  the  table,  but  the  index  (as  already  explained)  is 
obtained  by  the  rule:  One  less  than  the  number  of  figures  on 
the  left  side  of  the  decimal  point .  Therefore,  in  order  to 
memorize  and  explain  this  rule,  the  following  examples  are 
inserted : 


Number. 

Logarithm. 

Number. 

Logarithm. 

8236 
823.6 
82.36 
8.236 
0.8236 

3.915716 
2.915716 
1.915716 
0.915716 
9.915716  —  10 

0.08236 
0.008236 
0.0008236 
0.00008236 
0.000008236 

8.915716—10 
7.915716  —  10 
6.915716  —  10 
5.915716  —  10 
4.915716—  10 

Multiplying  or  dividing  a  number  by  any  power  of  10 
does  not  change  the  mantissa  in  the  corresponding  logarithm, 
but  only  the  index ;  for  instance  : 


LOGARITHMS.  73 

Log.  0.5  =  9.698970  —  10, 
and  Log.  500  =  2.698970,  etc. 

Thus,  the  mantissa  of  a  logarithm  is  the  same  whether  the 
number  is  0.5,  5,  50,  500,  5,000,  etc.  It  is  only  the  index  that 
is  changed  ;  therefore,  when  a  number  consists  of  three  or  less 
figures,  its  logarithm  is  found  in  the  tables  by  taking  the 
mantissa  found  in  the  first  column  to  the  right  of  the  number  ; 
that  is,  in  the  column  under  cipher.  The  index  is  found  by  the 
same  rule  as  before.  For  instance,  logarithm  to  537  will  be 
2.729974. 

To  Find  the  Logarithm  of  a  Number  Consisting  of 
Four  Figures. 

First  find  the  figures  in  the  column  headed  "  N  "  corre- 
sponding to  the  first  three  figures  of  the  number ;  in  line  with 
these  figures,  in  the  column  headed  by  the  fourth  figure,  will  be 
found  the  mantissa  of  the  logarithm  corresponding  to  the 
complete  number.  By  prefixing  the  index,  according  to  the 
rule  already  given,  the  complete  logarithm  is  obtained. 

EXAMPLE. 

Find  logarithm  of  5375. 

Solution : 

Under  the  heading  "  N  "  find  537 ;  and  in  the  column  at  the 
top  of  the  table  find  "5";  under  5  in  the  line  with  537  is 
730378. 

This  is  the  mantissa  of  the  logarithm.  The  index  for  a 
number  consisting  of  four  integers  is  3,  therefore  the  complete 
logarithm  of  5375  is  3.730378. 

To  Find  the  Logarithm  of  a  Number  Having  flore  Than 
Four  Figures. 

EXAMPLE  1. 

Find  the  logarithm  to  3658.2. 

Solution : 

Log.  3658  =  3.563244  and  log.  3659  =  3.563362  ;  therefore  the 
logarithm  for  3658.2  must  be  somewhere  between  the  two  logar- 
ithms thus  found  in  the  table.  The  difference  between  these 
two  logarithms  is  0.000118;  that  is,  if  the  number  is  increased 
by  1  the  logarithm  increases  0.000118,  therefore  if  this  number 
is  increased  0.2  the  corresponding  logarithm  must  increase  0.2 
times,  0.000118  =  0.0000236,  which  may  be  taken  as  0.000024. 

Thus : 

Log.  3658  =  3.563244 
Difference  corresponding  to  0.2  =  0.000024 

Log.  3658.2  =  3.563268 


74  LOGARITHMS. 

It  is  unnecessary  to  calculate  the  difference,  as  the  average 
difference  between  the  logarithms  in  each  line  is  given  in  the 
column  headed  "  i>"  in  the  tables.  The  difference  in  this  case 
is  given  in  the  table  as  119. 

EXAMPLE  2. 

Find  logarithm  to  1892.5. 

Solution  : 

The  mantissa  of  the  number  1892  is  given  in  the  table  as 
276921.  The  difference  is  given  as  229.  The  index  for  a  num- 
ber consisting  of  four  integers  is  3.  Thus : 

Log.  1892  —  3.276921 
0.5  X  0.000229  =  0.0001145  =  115 

Log.  1892.5  =  3.277036 
EXAMPLE  3. 
Find  logarithm  to  85673. 
Solution : 

The  mantissa  for  the  number  85670  is  given  in  the  table  as 
932829.  The  difference  is  given  in  the  table  as  51.  The  index 
for  a  number  consisting  of  five  integers  is  4. 

When  an  increase  of  10  in  the  number  increases  the  log- 
arithm 0.000051  an  increase  of  3  must  increase  the  correspond- 
ing logarithm  0.3  times  0.000051.  Thus: 

Log.  85670  =  4.932829 

0.3  X  0.000051  =  0.0000153   = 15 

Log.  85673  =  4.932844 

These  calculations  (or-  interpolations  as  they  are  usually 
called)  are  based  upon  the  principle  that  the  difference  between 
the  numbers  and  the  difference  between  their  corresponding 
logarithms  are  directly  proportional  to  each  other.  This,  how- 
ever, is  not  strictly  true ;  but  within  limits,  as  it  is  used  here, 
it  is  near  enough  for  practical  results. 

To  Find  the  Number  Corresponding  to  a  Given 
Logarithm. 

EXAMPLE  1. 

Find  the  number  corresponding  to  the  logarithm  2.610979. 

Solution : 

Always  remember  when  looking  for  the  number  not  to 
consider  the  index,  but  find  the  mantissa  610979  in  the  table. 
In  the  same  line  as  this  mantissa,  under  the  heading  "  N,"  is 
408,  and  on  the  top  of  the  table  in  the  same  column  as  this 
mantissa  is  3;  thus,  the  number  corresponding  to  this  mantissa 
is  4083  and  the  in4ex  of  the  logarithm  is  2  \  consequently  the 


LOGARITHMS.  75 

number  is  to  have  three  figures  on  the  left-hand  side  of  the 
decimal  point;  thus,  the  number  corresponding  to  the  logarithm 
2.610979  will  be  408.3. 

EXAMPLE  2. 

Find  the  number  corresponding  to  the  logarithm  3.883991. 

Solution : 

This  mantissa  is  not  in  the  table.  The  nearest  smaller 
mantissa  is  883945,  and  to  this  mantissa  corresponds  the  number 
7055.  The  nearest  larger  mantissa  is  884002,  and  to  this 
corresponds  the  number  7656. 

Thus,  an  increment  in  the  mantissa  of  57  increases  the 
number  by  1,  but  the  difference  between  the  mantissa  883945 
and  883991  is  46,  therefore  the  number  must  increase  if  =0.807. 

Number  of  Log.  3.883945  =  7655 

Difference  0.000046  =       0.807 

Number  of  Log.  3.883991  =  7655.807 


Addition  of  Logarithms. 

(MULTIPLICATION.) 

Where  the  logarithms  of  the  factors  have  positive  indexes, 
add  as  if  they  were  decimal  fractions,  and  the  sum  is  the  log- 
arithm corresponding  to  the  product. 
EXAMPLE  1. 

Multiply  81  by  65  by  means  of  logarithms. 
Solution : 

Log.  81  =  1.908485 
Log.  65  =  1.812913 

3.721398 

and  to  this  mantissa  corresponds  the  number  5265.  The  index 
is  3;  therefore  the  number  has  no  decimals,  as  it  consists  of  only 
four  figures. 

To  Add  Two  Logarithms   when  One  Has  a  Positive  and 
the  Other  a  Negative  Index. 

EXAMPLE  2. 

Multiply  0.58  by  32.6  by  means  of  logarithms. 

Solution ; 

Log.  0.58  =  9,763428  —  10 
,  38.8  as  1.513218 

11,270646  ~~10 


76  LOGARITHMS. 

This  reduces  to  1.276646  and  to  this  logarithm  corresponds 
the  number  18.908.  This  mantissa,  276646,  cannot  be  found  in  the 
table,  but  the  nearest  smaller  mantissa  is  276462,  and  the  differ- 
ence between  this  and  the  next  is  found  by  subtraction  to  be 
230,  and  the  difference  between  this  and  the  given  mantissa 
is  184. 

Thus: 

Given  logarithm  1.276646 
To  the  tabulated  log.  1.276462      corresponds     18.90 

Difference  0.000184      gives  _°i0_°i 

Thus,  logarithm  1.276646      gives  number  18J908 

To  Add  Two  Logarithms,  Both   Having  a  Negative 
Index. 

Add  both  logarithms  in  the  same  manner  as  decimal  frac- 
tions, and  afterwards  subtract  10  from  the  index  on  each  side 
of  the  mantissa. 

EXAMPLE. 

Multiply  0.82  by  0.082  by  means  of  logarithms. 

Solution : 

Log.  0.82    =  9.913814  —  10 

Log.  0.082  =  8.913814  —  10 

18.827628  —  20 

By  subtracting  10  on  each  side  of  the  mantissa  this  logar- 
ithm reduces  to  8.827628  —  10  and  to  the  mantissa  827628  corre- 
sponds the  number  6724,  but  the  negative  index  8 —  10 

indicates  that  this  first  figure  6  is  not  a  whole  number,  but  that 
it  is  six-hundredths  ;  therefore  a  cipher  must  be  placed  between 
this  6  and  the  decimal  point  in  order  to  give  6  the  right  value 
according  to  the  index;  thus,  to  the  logarithm  8.827628  —  10 
corresponds  the  number  0.06724. 

Subtraction  of  Logarithms. 

(DIVISION.) 
Logarithms  are  subtracted  as  common  decimal  fractions. 

To  Subtract  Two  Logarithms,  Both  Having  a  Positive 
Index. 

EXAMPLE. 

Divide  490  by  70  by  means  of  logarithms. 

Solution : 

Log.  490  =  2.690196 

Log.    70  —  1.845098 
0.845098 


LOGARITHMS.  77 

and  to  the  mantissa  of  this  logarithm  corresponds  the  number 
7  or  70  or  700  or  7000,  etc.,  in  the  table  of  logarithms,  but  the 
index  of  this  logarithm  is  a  cipher ;  therefore  the  answer  must 
be  a  number  consisting  of  one  figure,  thus  it  must  be  7. 

To  Subtract  a  Larger  Logarithm   From  a  Smaller  One. 

This  is  the  same  as  to  divide  a  smaller  number  by  a  larger 
one.  Before  the  subtraction  is  commenced  add  10  to  the  index 
of  the  smaller  logarithm  (that  is,  to  the  minuend)  and  place 
—  10  after  the  mantissa,  then  proceed  with  the  subtraction  as 
if  they  were  decimal  fractions. 

EXAMPLE. 

Divide  242  by  367  by  means  of  logarithms. 

Solution : 

Log.  242  =  2.383815  =  12.383815  —  10 

Log.  367  =  2.564666 

9.819149  —  10 

and  to  the  mantissa  of  this  logarithm  corresponds,  according 
to  the  table,  the  number  6594,  but  the  negative  index,  9  — 10, 
indicates  it  to  be  0.6594. 

Thus,  242  divided  by  367  =  0.6594. 

Multiplication  of  Logarithms. 

(INVOLUTION.) 

To  multiply  a  logarithm  is  the  same  as  to  raise  its  corre- 
sponding number  into  the  power  of  the  multiplier. 

Logarithms  having  a  positive  index  are  multiplied  the  same 
as  decimal  fractions.  Thus : 

Square  224  by  means  of  logarithms. 
Solution: 

2  X  log.  224  =  2  X  2.350248  =  4.700496  =  50176 

Logarithms  having  a  negative  index  are  multiplied  the 
same  as  decimal  fractions,  but  an  equal  number  is  subtracted 
from  both  the  positive  and  the  negative  parts  of  the  logarithm, 
in  order  to  bring  the  negative  part  of  the  index  to  —  10. 

EXAMPLE  1. 

Square  0.82  by  means  of  logarithms. 

Solution : 

2  X  log.  0.82  =  2  X  (9.913814  —  10)  =  19.827628  —  10,  and 
subtracting  10  from  both  the  positive  and  the  negative  parts  of 
the  logarithm,  the  result  is  9,827628  —  10 ;  this  gives  the  num- 
ber 0.6724, 


7  8  LOGARITHMS. 

EXAMPLE  2. 

Raise  0.9  to  the  1.41  power. 

Solution  : 

1.41  X  log.  0.9  =  1.41  X  (9.954243  —  10)  =  14.035483  —  14.1 

In  this  example  10  cannot  be  subtracted  from  both  parts 
of  the  logarithm,  but  4.1  must  be  subtracted  in  order  to  get  —  10, 
after  the  subtraction  is  performed.  The  logarithm  will  then 
read  9.935483  —  10,  which  corresponds  to  the  number  86195,  and 
the  negative  index,  9  —  10,  makes  this  0.86195. 

Division  of  Logarithms. 

(EVOLUTION.) 

To  divide  a  logarithm  is  the  same  as  to  extract  a  root  of 
the  number  corresponding  to  the  logarithm. 

Logarithms  having  a  positive  index  are  divided  the  same 
as  common  decimal  fractions. 

EXAMPLE. 

Extract  the  cube  root  of  512  by  means  of  logarithms. 
Solution  : 

**•  512    =  2.70927  = 

30 
O 

and  the  number  corresponding  to  this  logarithm  is  8,  80,  800, 
8,000,  etc.,  but  the  index  of  this  logarithm  is  a  cipher  ;  there- 
fore the  answer  must  be  a  number  consisting  of  one  integer, 
consequently  it  must  be  8. 

To  Divide  a  Logarithm  Having  a  Negative  Index. 

Select  and  add  such  a  number  to  the  index  as  will  give 
10  without  a  remainder  for  the  quotient  in  the  negative  index 
on  the  right-hand  side  of  the  mantissa  after  division  is  per- 
formed. 

EXAMPLE  1. 

Extract  the  square  root  of  0.64  by  means  of  logarithms. 


=  19^18-20 
2  2 

and  to  this  logarithm  corresponds  the  number  0.8". 
EXAMPLE  2. 
Extract  the  cube  root  of  0.125  by  means  of  logarithms. 


LOGARITHMS.  7  9 


Solution : 


log.  0.125  _  9.0961)1—10 _  29.09691—30 _  9  69g97  _  1Q 

33  3 

and  to  this  logarithm  corresponds  the  number  0.5. 

EXAMPLE  3. 

Extract  the  1.7  root  of  0.78. 

Solution : 

log.  0.78  9.892095  —  10 

1.7  1.7 

We  cannot  here,  as  in  previous  examples,  add  a  multiple  of 
10  to  the  index  on  each  side  of  the  mantissa,  but  7  must  be 
added  in  order  that  the  negative  quotient  shall  be  —  10  after 
the  division  is  performed.  Thus: 

9.892095-10       10.892005  -17 


1.7  1.7 

and  to  this  logarithm  corresponds  the  number  0.864. 

Short  Rules  for  Figuring  by  Logarithms. 

MULTIPLICATION. 

Add  the  logarithms  of  the  factors  and  the  sum  is  the  logar- 
ithm of  the  product. 

DIVISION. 

Subtract  divisor's  logarithm  from  the  logarithm  of  the  divi- 
dend and  the  difference  is  the  logarithm  of  the  quotient. 

INVOLUTION. 

Multiply  the  logarithm  of  the  root  by  the  exponent  of  the 
power  and  the  product  is  the  logarithm  of  the  power. 

EXAMPLE. 

Log.  862  =  2  X  log.  86  =  2  X  1.934498  =  3.868996 
and  to  this  logarithm  corresponds  the  number  7396. 

EVOLUTION. 

The  logarithm  of  the  number  or  quantity  under  the  radical 
sign  is  divided  by  the  index  of  the  root,  and  the  quotient  is  the 
logarithm  of  the  root. 


80  LOGARITHMS. 

EXAMPLE. 

4                    log.  2401          3.380392 
Log.  y  2401    =  -  "^j = | =  0.845098 

and  this  logarithm  corresponds  to  the  number  7. 

EXPONENTS. 

The  logarithm  of  a  power  divided  by  the  logarithm  of  the 
root  is  equal  to  the  exponent  of  the  power. 

EXAMPLE. 

8*  =  64 


x 


x  = 


log.S 
1.80618 
0.90309 

x  =  2 

The  logarithm  of  a  quantity  under  the  radical  sign  divided 
by  the  logarithm  of  the  root  is  equal  to  the  index  of  the  root. 

EXAMPLE.  x  _ 

8  =  \/512 

x  =     l°8'  512 

x  =  2.70927 
0.90309 
x  =3 

The  reason  for  these  last  rules  may  be  understood  by  re- 
ferring to  the  rules  for  Involution  and  Evolution  ;  for  instance  : 

862=  7396,  and  this  expressed  by  logarithms  is: 
2  X  log.  86  =  log.  7396. 

Therefore:     lo?'  7896  =   2. 


log.  86 


FRACTIONS. 


The  logarithm  of  a  common  fraction  is  found,  either  by  first 
reducing  the  fraction  to  a  decimal  fraction,  or  by  taking  the 
logarithm  of  the  numerator  and  the  logarithm  of  the  denominator 
and  subtracting  the  logarithm  of  the  denominator  from  the  log- 
arithm of  the  numerator;  the  difference  is  the  logarithm  of 
the  fraction, 


LOGARITHMS.  8 1 

EXAMPLE. 

Log.  %  =  log.  3  —  log.  4 

Log.  3  =  0.477121  =  10.477121  —  10 

Log.  4  —  0.602060 

Thus,  log.  %  =  9.875061  —  10 
This  is  also  the  logarithm  of  the  decimal  fraction  0.75. 


RECIPROCALS. 

Subtract  the  logarithm  of  the  number  from  log.  1,  which  is 
10.000000  —  10,  and  the  difference  is  the  logarithm  of  the  reci- 
procal. 

EXAMPLE. 

Find  the  reciprocal  of  315. 

Solution : 

Log.  1      =  10.DOOOOO  —  10 
Log.  315  —    2.498311 

Log.  reciprocal  of  315  =  7.501689  —  10 

To  this  logarithm  corresponds  the  decimal  fraction 
0.0031746,  which  is,  therefore,  the  reciprocal  of  315. 


Simple  Interest  by   Logarithms. 

Add  logarithm  of  principal,  logarithm  of  rate  of  interest, 
and  logarithm  of  number  of  years  ;  from  this  sum  subtract  log- 
arithm of  100.  The  difference  is  the  logarithm  of  the  interest. 

EXAMPLE. 

Find  the  interest  of  #800  at  4%  in  5  years. 

Solution  : 

Log.  800  =  2.90309 

Log.  4      =  0.60206 
0.69897 


40412 
Log.  100  =  2.00000 
Log.  interest  =  2.20412  =  $160  =  Interest. 


#2  LOGARITHMS. 

Compound  Interest  by  Logarithms. 

When  the  interest,  at  the  end  of  each  period  of  time,  is 
added  to  the  principal  the  amount  will  increase  at  a  constant 
rate;  and  this  rate  will  be  the  amount  of  one  dollar  invested 
for  one  period  of  the  time.  For  instance  :  If  the  periods  of  time 
be  one  year  each,  then  $30  in  3  years  at  5  %  compound  interest 
will  be  : 

$30  X  1.05  =  $31.50    at  the  end  of  first  year. 
$31.50  X  1.05  =  $33.075  at  the  end  of  second  year. 
$33.075  X  1.05  =  $34.73    at  the  end  of  third  year. 
This  calculation  may  be  written  : 

$30  X  1.05  X  1.05  X  1.05  =  $34.73 
which  also  may  be  written 

$30  X  (1.05)3  =  $34.73. 

Thus,  compound  interest  is  a  form  of  geometrical  progres- 
sion, and  may  be  calculated  by  the  following  formulas  : 

a  =p  X  r" 
Log,  a  ==•  n  X  log.  r  +  log.  p 


Log.  p  =  log.  a  —  n  X  log.  r 


log.  r 
Log.  r  =  !og. 


n 

p  =  Principal  invested. 
n  =  The  number  of  periods  of  time. 
a  =  The  amount  due  after  n  periods  of  time. 
r  =  The  amount  of  $1  invested  one  period  of  time. 

NOTE.  —  The  quantity  r  is  always  obtained  by  the  rule  : 

Divide  the  rate  of  interest  per  period  of  time  by  100,  and 
add  1  to  the  quotient. 

EXAMPLE. 

What  is  the  amount  of  $816  invested  6  years  at  4%  com- 
pound interest? 


LOGARITHMS.  83 

Solution  by  formula  : 

Log.  a  =  n  X  log.  r  -j-  log.  p 

Log.  a  =  6  X  log.  1.04  +  log.  816 

Log.  a  =  6  X  0.017033+  2.911690 

Log.  a  =  0.102198  +  2.911690 

Log.  a  =  3.013888 

a  =  $1032.49  =  Amount. 
EXAMPLE. 

If  $750  is  invested  at  3%   compound  interest,  how  many 
years  will  it  take  before  the  amount  will  be  $950. 

Solution  by  formula: 

n=     log-  a  — 


log.  r 

lo&-  95°  ~  l°£-  75° 
log.  1.03 

2.977724  - 


0.012837 
EXAMPLE. 

A  principal  of  $3750  is  to  be  invested  so  that  by  compound 
interest  it  will  amount  to  $5000  in  six  years.  Find  rate  of 
interest. 

Solution  by  formula  : 
Log.  r  = 

Lo     r  = 


n 
3-698970  —  3.574031 


6 
Log.  r  =  0.020823 

r  =  1.0491 

Rate  of  interest  =  lOOr— 100  =  100  X  1.0491  —  100  =  4.91%  ; 
or  5  %  per  year  (very  nearly). 

Discount  or  Rebate. 

When  calculating  discount  or  rebate,  which  is  a  deduction 
upon  money  paid  before  it  is  due,  use  formula: 


Log.  p  =  log.  a  —  n  X  log.  r 


84  LOGARITHMS. 

EXAMPLE. 

A  bill  of  $500  is  due  in  3  years.  How  much  cash  is  it  worth 
if  3%  compound  interest  should  be  deducted. 

Log.  p  =  log,  a  —  n  X  log.  r 
Log.p  =  2.698970  —  3  X  0.012837 
Log.p  =  2.698970  —  0.038511 
Log.p  =  2.660459 

p  =  $457.57  =  Cash  payment. 

NOTE.  —  Such  examples  may  be  checked  to  prevent  miscal- 
culations, by  multiplying  the  result  (the  cash  payment),  by  the 
tabular  number  given  for  corresponding  number  of  years  and 
percentage  of  interest  in  table  on  page  23  ;  if  calculations  are 
correct,  the  product  will  be  equal  to  the  original  bill.  For  in- 
stance, 457.57  X  1.092727  =  499.99909339  =  $500.00.  Thus,  the 
calculation  in  the  example  is  correct. 

Sinking  Funds  and  Savings. 

If  a  sum  of  money  denoted  by  £,  set  apart  or  saved  dur- 
ing each  period  of  time,  is  put  at  compound  interest  at  the  end 
of  each  period,  the  amount  will  be  : 

a  =  b  at  the  end  of  the  first  period. 

a  =  b  +  br  at  the  end  of  the  second  period. 

a  =  b  +  br  +  bi*  at  the  end  of  the  third  period. 

At  the  end  of  n  periods  the  last  term  in  this  geometrical 
series  is  brn~l  and  the  first  term  is  £,  while  the  ratio  is  r. 
The  sum  of  the  series  is  the  amount  which  according  to  the 
rules  for  geometrical  progression  (see  page  69)  will  be  : 


r—  1 


r  —  1 

EXAMPLE. 

At  the  end  of  his  first  year's  business  a  man  sets  apart 
$1200  for  a  sinking  fund,  which  he  invests  at  4%  per  year.  At 
the  end  of  each  succeeding  year  he  sets  apart  $1200  which  is 
invested  at  the  same  rate.  What  is  the  value  of  the  sinking 
fund  after  7  years  of  business  ? 
Solution  : 

=  1200  X  (1.047  —  1) 

1.04  —  1 
a  =  1200  X  0.31593 

0.04 
a  =  $9477.90 


LOGARITHMS.  85 


EXAMPLE. 


A  man  20  years  old  commences  to  save  25  cents  every 
working  day,  and  places  this  in  a  savings  bank  at  4%  interest, 
computed  semi-annually.  How  much  will  he  have  in  the  bank 
when  he  is  36  years  old?  (NOTE.—  25c.  a  day  =  #1.50  a 
week  =  26  X  $1.50  =  $39  in  six  months.  4  %  per  year  =  2  % 
per  period  of  time  ;  36  —  20  =  16  =  32  periods  of  time). 

Solution  by  formula: 


r  —  1 

=  39  X  (1.0232  —  1) 
1.02  —  1 

=  39  X  (1.8845  —  1) 
0.02 

a  =  39  X  0.8845  X  50 
a  =  1724.775  =  $1724.77  =  Amount. 
Thus,  in  16  years  a  saving  of  25c.  a  day  amounts  to  $1724.77. 

If  the  money  is  paid  in  advance  of  the  first  period  of  time 
the  terms  will  be  : 

a  =  br  at  the  end  of  the  first  period. 

a  =  br  +  br*  at  the  end  of  the  second  period. 

a  =  br  +  br*  -\-  br*  at  the  end  of  the  third  period. 

At  the  end  of  n  years  the  last  term  in  this  geometrical 
series  is  br11  and  the  first  term  is  br,  while  the  ratio  is  r.  The 
sum  of  the  series  is  the  amount,  which,  according  to  rules  for 
geometrical  progressions  (see  page  69),  will  be  : 

r     r°    ~~  br 


r  —  1 

_  br   (rn  _  1} 
r  —   1 

EXAMPLE. 

Assume  that  the  man  mentioned  in  previous  example, 
instead  of  commencing  to  save  money  when  20  years  old, 
already  had  $39  to  put  in  the  bank  at  4  %  the  first  period  of 
time,  and  that  he  always  kept  up  paying  $39  in  advance  semi- 
annually.  How  much  money  would  he  then  save  in  16  years  ? 


86  LOGARITHMS. 

Solution  : 

=  39  X  1.02  X  (10232  —  1) 

1.02  —  1 
_  39  X  1.02  X  0.8845 

~002- 
a  =  1759.27 

Thus,  by  paying  the  money  in  advance  semi-annually,  he 
will  gain  (1759.27  —  1724.77)  =  #34.50. 

If  a  principal  denoted  by  P  is  invested  at  a  given  rate  of 
compound  interest,  and  successive  smaller  or  larger  equal  pay- 
ments denoted  by  b  are  made  at  the  end  of  each  period  of  time 
so  that  they  will  commence  to  draw  interest  at  the  beginning  of 
the  following  period  at  the  same  rate  as  the  principal,  the 
formula  will  be  : 


but  for  logarithmic  calculations  it  is  more  convenient  to  denote 
the  rate  of  interest  by  y  %  and  the  formula  will  read  : 


a  =  p   X    rn     +          ~J~ 

100 

a  =  jxr*  +  100*(r»-l) 

y 

b  _  ay—py  X  rn 
100  rn  —  100 

rn=.    a.y  +  100* 
py  +  100  6 

4-  100  b 


=  log.  r 

lo?  *  y  +  10°  * 

*'    p  y  +  100  ^ 
-    -          -S- 

NOTE.  —  Using  these  formulas  it  must  be  understood  that 
«  represents  the  number  of  periods  of  time  that  the  principal  is 
invested,  and  that  this  first  period  is  considered  to  be  the  period 
at  the  end  of  which  the  first  payment,  b,  is  made. 

EXAMPLE. 

A  man  has  $50  in  a  savings  bank  and  he  also  puts  in  $25 
every  month,  which  goes  on  interest  every  6  months  ;  the  bank 
pays  4%  interest,  computed  semi-annually.  How  much  money 


LOGARITHMS.  &7 

can  he  save  in  5  years  in  this  way?      (NOTE.— 4%  per  year  = 
2%  per  6  months,  or  per  period  of  time,  and  $25  a  month  =  $150 
every  6  months,  or  per  period  of  time.    The  interest  is  computed 
semi-annually  ;  therefore  5  years  =  10  periods  of  time). 
Solution  by  formula : 


y 

a  =  50  X  1.02-  +  100  X  150  X  (102 "-I) 

a  =  50  X  1.219  +  100  X  150  X  0.219 

2 

a  =  60.95  +  1642.50 

a  =  $1703.45  =  Amount. 

The  original  sum  of  $50  has  increased  to  $60.95,  and  the 
monthly  payments  amounted  to  $1500.  The  last  six  payments 
did  not  draw  any  interest,  as  they  were  deposited  in  the  last 
six  months  of  the  fifth  year  and  would  commence  to  draw  inter- 
est at  the  beginning  of  the  sixth  year  if  the  amount  had  not 
been  withdrawn. 

EXAMPLE. 

A  man  has  $800  invested  at  5%.  How  much  must  he  save 
and  invest  at  the  same  interest  every  year  in  order  to  increase  it 
to  $3000  in  five  years  ?  Interest  is  computed  annually. 

Solution  by  formula : 

b  _  ay—  py  X  r* 

100  r*  —100 
b  __  3000  X  5  —  800  X  5  X  1.05s 

100  X  1.055  —  100 
b  _  15000  —  1.2763  X  4000 

100  X  1.2763  —  100 
15000  —  5105.2 


b  — 


127.63  —  100 

9894.8 


27.63 

b  =  358.118  =  $358.12  to  be  paid  in  each  year. 
The  total  payments  will  be : 

800  +  5  X  358.12  =  $800  +  $1790.60  =  $2590.60. 
The  rest  of  the  amount  is  accumulated  interest.     The  last  pay- 
ment is  made  at  the  end  of  the  fifth  year ;  therefore  this  money 
does  not  draw  interest. 


88  LOGARITHMS. 

EXAMPLE. 

A  man  calculates  that  if  he  had  $1800  he  would  start 
in  business.  He  has  only  $120,  but  is  earning  $15  a  week  and 
figures  that  he  can  save  half  of  his  weekly  earnings.  He  puts 
his  money  in  a  savings  bank,  where  it  goes  on  interest  every 
six  months,  at  the  rate  of  4%  a  year.  How  many  years  will  it 
take  him  to  save  the  required  amount?  (NOTE. — $7.50  a  week 
==  26  X  7>£  =  $195  in  six  months,  and  4%  per  year  =  2%  per 
six  months,  or  per  period  of  time). 

Solution  by  formula : 

log. 
n  =  


log. 


log,  r 

1800X2  +  100X195 


_      120  X  2  +  100  X  195 

log.  1.02 

,    3600  +  19500 
l°S'   240  +  19500 

log.  1.02 


One  period  =  6  months  ;  8  periods  =  4  years  ;  therefore, 
under  these  conditions  it  takes  four  years  to  save  this  amount 
of  money. 

If  a  certain  sum  of  money  is  withdrawn  instead  of  added, 
at  the  end  of  each  period  of  time,  the  formula  on  page  86  will 
change  to  : 


y 

Every  letter  denotes  the  same  value  as  it  had  in  the  formula 
on  page  86,  except  that  b  represents  the  sum  withdrawn  instead 
of  the  sum  added. 

EXAMPLE. 

A  man  has  $5000  invested  at  5%  interest  compounded  an- 
nually, but  at  the  end  of  each  year  he  withdraws  $200.      How 
much  money  has  he  left  after  six  years  ? 
Solution  : 

a  =  1.05"  X  5000  -  100  X  200  X  (1.05*  -  1) 

5 

a  =  1.34  X  5000  -  10°  X  ™  X  °'34 

a  =  6700  —  1360 

a  =  $5340  =  Amount. 


LOGARITHMS.  89 

If  the  deducted  sum,  £,  exceeds  the  interest  due  at  the  first 
period  of  time,  the  amount  a  will  become  smaller  than  the  prin- 
cipal/, and  in  time  the  whole  principal  will  be  used  up.  This 
will  be  when : 

100£(r«—  1) 
P  X  r»   =  -      -y- 

This  transposes  to 

100  £ 
= 


log. 


-py 

1003 

100  b    _  py 


log.r 

EXAMPLE. 

A  principal  of  $5000  is  invested  at  4%  per  year,  but  at  the 
end  of  each  year  $600  is  withdrawn.  How  long  will  it  take  to 
use  the  whole  principal  ? 

100  X  600 
_    l°S-    100  x  600  —  5000  X  4 

'log.  1.04 

60000 


60000  —  20000 


n  = 


log.  1.04 
log.  1.50 
log.  1.04 
0.176091 


0.017033 
n  =  10.3  years. 

Paying  a  Debt  by  Instalments. 

This  same  formula  applies  also  in  this  case;  for  instance: 
A  man  uses  $1500  every  year  toward  paying  a  debt  of  $10,000, 
and  5%  interest  per  year.  How  long  will  it  take  to  pay  it? 

100  X  1500 


'    100  X  1500  —  10000  X  5 

n  =  — 


log.  1.05 

150000 
100000 


n= 


log.  1.05 

log*  1.5 
log.  1.05 

=.    0.176091 

0.021189 

n  =  8.3  years, 


9o 


LOGARITHMS. 


(M  00  ^  O  SO  'N 

CO  <M  M  <M  -H  rH 


-H  O  to  CO  O  JO  CO  Ci  to   CO  O  CO  CO  O  i—  >O  n  Ci  to 

o  o  Ci  Ci    Ci  co  GO  i~  i-    i—  r-.  to  so  so     o  10  o  TJ<  Tt< 

T^  T}H  CO  CO      CO  CO  CO  CO  CO      CO  CO  CO  CO  CO      CO  CO  CO  CO  CO 


CM  rH   CM  tO 

tO  1—  SO  CM 


Ci  rH  O  Ci  t-  O   T*  tO  O  i-  CO   CO  T*  O  CO  CO   iH  CO  rH  T^  i-   rH  10  O  SO  10 

CM  7^  i—  t-  -*  t—      CO  rH  CO  O  Tfl      iO  <N  CO  SO  CO      t—  t-  O  Ci  O      Ci  ~t  1—  CO  CO 
O  CO  iO  i—  Ci  O      rH  7<1  CM  C-l  rH      O  Ci  1—  >O  CO      O  t—  T^  O  t-      <>\  CO  CO  CO  CO 


•H  »O  Ci  "*      CO  'M  SO 

SrHrHrHT-l       C^COCO"^Tfl 
OOOO     OOOOO 


t-  ^h  O  t 
O  Tt<  CO  rH 


Ci  ci  c: 


CO^t^O'N-*     J—  MOrHiO 

CiCiT^SOCOtO        OrHCOrHlO) 

xOCOrHCOiOSO  J-COCOCOt— 
rH  IO  Ci  CO 
CO  CO  CO  Tfl 


T^  CO  CO  >O  C 
SO  CO  1-  CO  i 

SO  O  CO  rH  Ci   O  rf 

t-  TH  10  Ci  cq  so  o 
o'o'SSo1 


•^lOrHt-  ROCOCO 

O   CO   CO  TH  CO   Ci    rH    rH 

O  J^~  CO  Ci  ~^  O  O 

T+H  t—  rH  Tf  CO   'M   O 

r^  t^  i'*  co  co  co  Ci  Ci 


SO  7-1      OCOO^CN      lOCOSO 


rHTtl^-CirH   ? 

fM  SO  O  T*  OS  C/ 


•3 


Ci  c 

C 


O  CO  SO  >O  >'l      CO  X  00  CO  1—     O  O  I—  OS  rt< 

.—  OCiOCO      .-MCOrHSOCO      I—  CO  SO  to  ^t 
M  rH  Ci   X   »O       CO  O  1—  CO  O       »O  rH  tO  rH  SO 


TflXOrHOrH  C<|-i<Ol^-Ci  lOSOCOSOO  COXJM-^t1 

t—  O  CO  iO  t—  X  CiOOOCi  X  t—  O  Tfi  vN  CitOCOO 

81"1   O   rH    — I   rH    rl  C^l    CO   CO   CO   TjH  "*    >O    lO   >O   SO  O   tO  f^  t~ 

ooooo  ooooo  ooooo  oooo 


CirH»T5OCO 
S<lCi 

CO 

r       Tf 

Ci   Ci 

oo 


Ci 

(Mt-COCOCN 

-^  t—   rH    Tf 


rH   Ci   tO   O  T^   CO 

-*-%%% 
^2^ 

000 


OCOOOSO  rH  tO  CO  rH  — I 

COCiGO^iSO  SOC^tOi—  O 

iO(MCiSCfN  CO-tCi-tC: 

Ci  (N  to  O  CO  t—  O  -t  1— 

8t^t—  co  coX)CiC:c: 

ooo  ooooo 


OOrHrHOXtO       lO>Ot~COCN       lOCOSOSOCM 

SOXOXSOrH      CMCSCNCNX      O  OS  TJI  CO  iO 
O  O  O  -H  I-H  7^      O-l  CO  CO  CO  "^      -HH  -HH  iO  i.O  SO 


»0  CS  SO  iO  h-     -fS 

rHCO(N'MGO       rH 


'OOOOO     OOOOO     OOOOO 


•^  SO  CO      CO  X  X  iO  X 

01   Ci        CM    rH   1-   O   Ci 

"CO      CO  7-1  O  Ci  tO 
-H      O  Ci  CO  SO  O 

-t  -t  10  10  so 
ooooo 


tO  CM  I—  -H  OOiOONO 

COOOTfCO  GOtOOCNrH 

rHCOlOrH  I—  CO  Ci  Tf  Ci 

COrHiOCi  <M  SO  CS  CO  SO 

1^  I—  1-  CO  CO  CO  Ci  Ci 

ooooo 


000      O 


§HN«^i!5      OJ-OOOO      H«CC^»«      OI'WCSO 
OOOOO       OOOOH       rHiHrHrHH      HHrHHN 


LOGARITHMS. 


CO  TH 
CO  CO 


O  GO  ire  CO  TH         GO  CO  -t<  TH 


>O  CO  ^H 


£8 


1 
Ssss 


TH  CO  Ci    CO  HH  -M  CO 
CO  TH  TH    O  CO  O  TH 

t-003     fo^oo 


TH   <N  73  TH  CO 


O  »O  CO  CO 


ire  CM  Ci 

Ci  O  CO 
CO  CO'  CC 

co  ire  co 
t—  I—  I— 


1—  TH  Ci  O  CO 
I—  Ci  t 1  t- 


-H  — (  GO  O         Ci  T*  CO  CO  H< 

Ci  CO  CO  Ci  OO       ire  CO  CO  ire  GO 

TH  00  co  co  co      co  co  -t  -f  -t 


O  O  TH  CO  CO 
CO  GO  TH  7^  TH 
-t  CO  CO  (N  TH 


O  O  O  4-i  ^ 


o  ^  S  o  g     « 


II 


TH  -t  t- 

>o  >o  »o 


HJ<  co  Ci  sq  »re 

CO  CO  TH  CO  CO 

TH  O  O  Ci  CO 

co  Ci  n  "t  r- 

CO  CO  t—  1^-  i- 


THOCOOr-l 

TH    O    t—    TH 

OCi  (M  CO 


00— I  -N  05  COCOO^O 

O<  i— I  OO  3^  CO  COOO1—  t— 

O5  -N  -t  1^  O5  -r-t  CO  -t  »O  O 

CO  "N  »O  CO  T— (  »O  OO  i— I  "*  t— 

TH  n  -M  c<i  co  co  co  ~t  't  ^ 


g^ 

o  o  »re  o  co 


TH  CO 

-^  Ci  co  ire  Tt 

i»  -*  -<ti  CO  <N 

»re  oo  -H  H^  r-- 
o  co  i—  i-  t— 


ire  o  o  co  GO     COTH<MTHCO 
co  co  co  i'*  Ci     ci  i"**  co  ire  ire 

iHCO'TOCOCO        COCO-fH/lTfl 


co  ire  co  co  co 

TH  OO  CO  CO  CO 

O  CO  CO  Ci  CO 

ire  >re  >re  ire  co 


H^  CO  TH  O  O5 

HH  o  ^  o  ire 

CO  CO  TH  O  Ci 

o  co  TH  H-  co 
o  co  i-  i-  i- 


O    -t   1-  O   HjH 

O     O     O     T-H     TH 


TH  -f  o  o  c<; 

1-  O  TH  Ci  ^ 


;M  >o 

t—  O 


TH  H^  -M  ire  HJ<  ci  T-I  ci  »re  co 

i—  i—  o  o  co  co  co  i—  TH  co 

oo  ire  oo  TH  co  »re  t-  co  o  TH 

t-  O  CO  t-  O  CO  CO  O  CO  CO 

TH  01  co  co  co  co  co  co  ~*  -<n 


t-  CO  CO  TH 

_  ts  cfj  Oil 
COCOTHO 
t—  O  CO  CO 

co  i-  i- 1- 


o      H 
^*      ^t 


LOGARITHMS. 


•  CO  "*  <N   TH  CS  CO  CO  »O   CO  <M  O  CS  < 
1  CD  CD  CD   CD  iO  *O  lO  lO   iO  lO  iO  "^  ' 

ICMCSCS     <M  CM  cs  cs  CM     cs  CM  CM  CM 


CO  CM  rt<  GO  CD  1-  TH  CS  TH  CO    CDTH 

CO  CO  rH  GO  »O  CM  OS  »O  CM 

OGOrHCOCD  CSiHT^t- 

OSCSOOO  O  TH  TH  rH  rH 

TH  TH  CM  CM  CM  cs  <M  CM  cs  CM 


CN  (N  CM  (N 


_       -^^  -H^  £*•    Q5 

rHrHrHrH  (M(M(MCSCO          COCOCO 

CM<M(M<M       CSCSCSCSCq       <M<M(M 


COOCOCOO        -^  T-I  O>  T-t  »O 
TH  1—  O  C<l  CO       THOOfMCDt— 

i-H  OS  t- 


cococo     -^  10  o  o  TJ< 

rH  CD  OS         rH  rH  <O  1^—  CM 
lOTHI--        rt<OCDTH£- 
TH  Tj<  CD  OS 


COCDOOTH        Tt^t-OCSlC        GOOCOCDQO 

GOGOGOCS         OSCSOOO         OTHrHrHrH         CSCMCMCMCO 

3^  (M 


^  CO  OS  TH 


co  o  >o  r—  TH 
<M  oo  c<<  rj<  »o 

rH  OS  OO  CD  •<* 
O  (N  iO  CO  TH 
GO  CO  GO  CC  OS 


CO  tO  CM  CO  O       TH  t-  OS  CD  • 


OS  C^  (M  CD  TH  OS  OS  rH  CD  "^ 

CO  O  ^  CD  1:-  lO  <M  GO  rH  CO 

GOt-lOCOi-l  CSt— rJHCSOS 

CS  :N  iO  CO  rH  CO  CD 

IT-  GO  GO  00  OS  CS  CS 


CO  CO 

CO  IT- 


<M  CD  ^  t-  »O 

OS  OS  GO  O  rH 

CO  <M 


§OS  CD  (M 
(M  »O  GO       O  CO  CD  CO  rH 
OO         O  rH  rH  — I  rH         CSCScMCMCO 


»O  GO  OS  OS  CO 


lOrHOCOO  1-IOCD 

OO3<«t-TH  COCOCN 

CO  O  1^-  CO  O  CD  CM  GO 

_...,       ir-  O  (M  »O  CO  OCO»O 

OSOO         OTHTHrH-rH  <M(MCMCMCO 

i-HCMCM       CMCMCMCMCM  S 


^  os  >a 
co  <M  r- 

<M  rH  OS 

t-  00  OO  GO  OS       OS  OS  OS  O 
THTHrHrHTH        THi—lTHCS 


iT"*  "^  TH  rH  (M  lO  O  1^—  JT"~  C^ 

CSGOCDOCO  rHCSCDCOrH 

COrHTt^t-O  COiOCO-rH^ 

t-  GO  GO  OO  CS  CS  CS  CS  O  O 

THTHrHrHTH  THi—lTHCSCM 


CM  -#  t-    O 
rH  TH  TH    CM 

sq  <N  CM  <M 


to  0^  i^*  o^  co  QO  ' 

i-(  O  OO  ^   O  <M  ' 
1^*  CO  OO  '^   O^  iO  ( 


T-I  «  n  ^  «s 


LOGARITHMS. 


93 


CO  l.O  CO  (N  rH        OS  X  t>  »O 
GN  <N  :N  <M  GN       S3S3^5^ 


X  CO  O  rH  OS  CO  ~f  'N  t-  OS 

-M  i-  rH  co  co  co  rH  x  co  i- 

t-  rH  co  O  -f  x  ?»  «-^  os  n 

i—  o  o^  »o  r-  a  'M  -f  -^  ci 


Q  7?  CO  <M  C 

CO  OS  ^4  »c  X 

rH  CO  CO   X  C 

t-  t-  1—  t-  X 

CM  3<J  -M  S<1  ;M 


12^^^^ 

OCOOXC 


18 


CO  »O  CO  O 

^  co  x  o 
1111 


S&|ss 
sll|ii 

^  <M  'M  n  -N 


"^  CO  CO  <N  CO 
OS  t-  ^  O  TJ< 
>f5  OS  CO  t-  O 

g  g  3  §  § 

r"i  -M  ?i  oq  GN 


:i§g 


i^ 

CO  CO  O  ^1  «O 

rH   CO  CO   X   O 

i—  r-  i—  t-  x 

GVJ  (M  CM  (M  SN 


^o 


COX  CN3 
IT-  OS  ^  ^f 


SS££S   S 


§  ss 


rH  CO  »O  X  O 
^^^^^ 


CM  O  CO  rH 
.  .  X  CO  CO  OS 

CO  X  rH  CO  O 

CM    Tft  t—  I 


'•N    '*N    1"^   W*    ^^ 

O  CN  Tt  «t  i— 

00000 
<M  (N  GM  CO  CO 

!*? 


UN   ^r1  i      ww  T™n         v«J  «u  ' 

^i^i^^    ^?5 


rH  CO   rH  CO  X 


fN  -N  (N  3<1 


CO   rH   CO    <M   X 
rH  O   t-  CO  1^ 

rH  lO  X  (M  O 

O  rH  CO  '^O  X 


^ 


3? 


O  CO  »O  £•  < 

S^^^^ 


t-    O    rH 

X  t-  CO 

t-    OS    rH 
»O 


^  §5ii 


iO  X  X  Tf  1^ 
-*  O  CO  CO  1- 

t~  rH   CO  O  Jt 

2S  •«*  »o  S  o 

(M  CS  C^  "N  (N 


1—  CO  CO  CO  -^ 

t-  CO  CO  O  -* 
X  GN  CO  O  CO 

X   rH   CO  >C  X 


t-  O  rH 

1  Q 


i-^  CO  O  X 
O  CO  »0  i-  O 

t- 1-  t-  1—  t- 

(M  G<1  «N  >?1  <N 


?5 


'  »O  O  rH  O 

i  so  O  CO  CO 
(N  GV|  GX|  <M 


^r  i.~  '~jj  uu 

I^§^2: 
££££ 


t-  'N  id 
"^  Tjn  3<1 

CO  >O  t- 


-M   O    rH   O   CO 

»C  O  i^t1  X  O 
(7<l  t^  rH  >O  O 

GN  ^  ^5  ^  ^4 


^^^ 


. 
(N  »-O  GO  TH  -Tt 

OtM-^t^os 

^(MiM^CN 


(M 


Hiii 


O3 


CO  •*  X  X  >O 


00  >O  O  1-1 
)-t<NOSTj< 


OOOi-H 
GCOtM(M 

cs  co 


-      - 

o  os     r-f  co 


OS  t—  ^ 
OSOO 

o  os  1-1  < 


O  O  "*  CO  -t       X  O 


^^^^^ 
S^^i 


00  t-  -^  I-H  I- 
t-  X  X  t—  Tt 
•^  i®  X  O  (M 
<M  •«*  O  OS  T-I 


COCO 
i-(  1— 

i?3  OS  " 


CO  OSi-li-lX-M 

t-  I—  t^OrHt- 

(7^1  CO  O  "^  X  T—  < 

lO  1^*  O  O^l  "^  1^* 

iO  i^iCOCOCOCO 

(N  Cq<NJN?N(N 


Or5^CCX 

CO  tr^  t**  t^*  1s-* 
G<l  (M  <N  (7^  G^l 


^ 


.^ 

w  X  rH 

OS  OS  O 
<M  <N  CO 


OJT'WCiO      TH«M^»a      cr-ac®o      IHNM^* 

*         *** 


94 


LOGARITHMS. 


CM   T-H  O  OS  X  b-   «O  »O  -t<  CO  CO 

O   O  O  OS  OS  OS   OS  OS  OS  OS  OS 

CM    CN  CM  T-l  T-H  TH    T-H  T-l  T-H  T-H  T-H 


CM 

^  —  ^  .1  CM   CM  CM  CO  CO  •: 

COCOCOCOCO   CO  CO  CO  CO  C 


T-HCMCOCO't1 

>O  t-  OS  T-H  CO       »O  1—  OS  I-H 

CM(N'NCO 

cococococo     cocococo 


OS 
CO 


<M     X»OCOCOT^ 

rH       b-COX(NO 
CM       i-Hi-iOO 


i-iOOOS 

l^OST-nO) 


,,.. 
COCOCOCOCO        COCOCOCOCO 


££i28 

CM  CM  CM  CO  CO   CO  CO  CO  -*  Tfl   -^  Tj<  Tf  i<7  »r? 

CO  CO  CO  CO  CO   CO  CO  CO  CO  CO   CO  CO  CO  CO  CO 


CO  tO  b-  O  CM 

CO  CO  CO  CO  CO 


OS  O1  — H  CN]  CO  "^  >O  b*  OS  CM       CO 

•^COXOCM  ^tCOXOCM       ^* 

T-Hi—iT-lCMCM  'CMCMCMCOCO 

COCOCOCOCO  COCOCOCOCO 


LOGARITHMS. 


95 


©CiX       X  F-  CO  »O  ^t1        Tf  CO  7^  i— i  i-H        © 
Ci  X  X       XX  XXX       XX  XXX       X 


CO  -t  -t  Ct  "M 
X  I-  ~  .-  -f 
O  I~  OS  -- 

cc  co  co  cc 

Of  -t  :O  O  C 
-t  O  O  to 
:r  O  -t  CC 

o  i-  c:  — 


i—1  Ci  O  CO  X  >(*"  '•"t4  CO  ©  1^*  CO  X  CM  Ci  CS  r-*  F*  ^  CO  ™H 

©  CC  CC  X  1^  CO  Ci  1-1  CO  CO  CO  tN  I-H  X  lO  -Nl—  C-ICO© 

CC  ^  ©  X  I-  lO  CO  CI  ©  X  CO  ^  (M  Ci  I-  id  0-1  ©  1-  O 

I~  Ci  ©  O}  ^f  CO  X  ©  I-H 


F- 

XXX 
CO  CO  CO 


?O  Ci  ^  i-H 
»O  Tf  CO  r-H 


•<t  01 

^H       CO  lO  t- 
l-  I-  1-  t~  X      X 

CO  CO  CO  CO  CO       CO 


XX 
CO  CO 


»C  i— i  O  <M  'fj 

•*v  o  >o  c;  'M 

co  co  ^o  co  co 


x  x  x  r~ 

_  ci  x  r~  co 
o  co  x  ©  c-i 


I-H  co  r-  co  I-H 

O  "N  X  -^  i " 

>O  -t  C-l  *-*  • 


i-HTfXCO>O  F-<MOiCii-l  F-OCO^X 

CO  CD  X  ©  i— i  i— ii— (Cll—  O  ^^l-C^t— © 

CO»Ot—  OSi-H  GM^COX©  T^i^i5J^T'^) 

COCCCOCOCO  COCOCOCOCO  COCOCOCOCO 


Siss 


t^  't  i-H   t-  CM 

i-l©CSF-CO  Th  (N  i-H  OS 

" '  1—  OS  i-(  CO  »O  t-  X 

CC  CC  CC  CC  CC  CO  CC  CO  CO 


OS  OS  Ci  X  CO  ^ 

CO  ^H  Ci  I—  &  CO 

©  d  CO  O  t—  Ci 

1—  t-  t—  I—  X  X  X  X  X  X 

CO        CO  CO  CO  CO  CO  CO  CO  CO  CO  CO 


X  CO  iG>  O  X 
©  -M  CO  CO  Cl 

^  0  r^  cl  l^ 

>-c  o  o  o  co 

CO  CO  CO 


-H  X  O  i— 'CO         i-H-^t— Oil— ( 

CC  -t  CO  "M  ©       Ci  I—  O  CO  CI       ©  X  CO  CC  — < 
COOl-Cii-l       (MTt^COX©       -MCOOt-Ci 


96 


LOGARITHMS. 


COOSCOOO  TH  rH  CM 

CN|  -*  CO  t-  t-  CO  O  CO    _ 

CMOSCDCOO  t~  "*  rH  X  "^       TH  t—  -tf  C 

COCO  OS  rH  CO  ^  CO    CO  CS  rH  C 


rH  CO  ^  CO    X  CS  rH  CO  -*    CO  t-  CS  —  ? 
TH  TH  rH  rH    TH  rH  -N  CM  CM    CM  CM  ?M  CO  C< 


lO  t-  OS  Ci  O 

^HCS 


i-H  CO  CS  CO  i-l 

TH    CO    TjH    U3    ^ 

I-  -*  i— I  CO  O 


t-  CO  CS  10  ^ 
O  -t1  CM  O  L— 

CM  OS  CO  CO  OS 


CO  O  CO 
CO  'CO  O  CM  CM 
CO  HH  — ,  t—  CO 
id  lr-  CS  O  CM 
CM  CM  CM  CO 


O  O  ^  I-H  Ol 
CO  -M  r-H  O  I- 
O>  O  I-H  t-  CM 

CO  >O  1—  CO  O 


CO      CO  CO  CO 


CO  T-H  GO  CO  — H 
CO  O  I—  CO  CS 
CM  CS  CO  CO 
CM  CO  id 


CM  -f  iO  CO 

CO  OS  JO  TH 


OS  t—      O  1—  1—  CM  CM 


1—  CO  OS  lO  1-1 

CO  >^  ^  CO  O 
CO  CO  CO  CO  -t 


CO  CS  CO  O  — i 
TH  O  CS  r-  "* 
CS  CO  CM  CS  CO 
"  CM  CO  iO 


8Tt<  CO  rH   rH 
CO    rH   CO   O 

CO  CS  CO  CM  CS 
i-  CO  O  CM  CO 
CNI  CM 


^  rH  CO  CO  Ir- 
CO  O  00  CS  O 
O  rH  1—  CO  O 
O  t—  CO  O  CM 
CM  CM  CM  CO  CO 


O  t-  CS  rf  -t 

TH  O  -OS  X  CO 
I-  CO  CS 

f-.  r/-.  r^ 


t—  nOCO-H 


w^  v  w  ^^> 

i—t  CO  lO  IT— 
00 


h    rH    CO   ^  TH   t- 

-     TH   TH    22   °          ^  °° 

•^   ^          TjH    ^   ^   ^   ^          ^^ 


T-HCOCO      O 
CMCO^      »O 


COCSCDCMX  ^tOCOCMX 

lOCOXOrH  COlOCOXCS 

(N  CM  CM  CO  CO  COCOCO 
•H^   ^   "H^  -H^  *H^ 


COCOCSiOO  OCO^t— X 

-            O  CO  O  t—  GO  OS  CO  GO  CO  """t1 

COGOOi-HCO        OCOCOOi— I  CO  Tji  CO  CO  CS 

TH  T-H  CM  CM  CM        'CN|  CM  CM  CO  CO  CO  CO  CO  CO  Ct 

^TjH^Tfl^ 


I—  rH  CO  O         »O  Tf<  t-  ^  iffl       OCSCMCS'rH 
COCMCOi—l        Th  IT—  OS  rH  CM      COCMCMOOS 
tr—  CO  OS  "^ 

CO  CO  CO  CO 
•^  ^  ^  "^ 


OCOOOCO         rHrHOTtCO         CMrHlO'CM^ 

^COCMOt-       ^*O»OO"^       XrHCOiOCi 

cocoes      COCOOSCOCM      co  »o  — t  r—  a^    w^  ^  ^  ^    -.-. 

COXOSrHCO         'tfCOCOCSrH       (M^COf-CS 
THrHrHCMrN       CMCMCMCMCO      COCOCOCOCO 

f   "^   ^   Tt<   "* 


O  ***  00 

to  ia  ia 

«   *?   N 


LOGARITHMS. 


97 


>O  CO  ift  <M  Tt 

~  CO  CC  OS  r-l 

i— I  CO  Tf<  CO 


lO  O  T-I  CD  CD  <M 

"*  TH  1—  <M  1—  <M 

i-H  CO  O  tf3  Ci  Tt 

rH  C^  "^  O  CO  GO 

t^ 

-t 


o  I-H  cq  t- 

00  t-  Hi  *M 

O  »O  O  >O 


GO   "<t  O  —  <M  X 
Ci   CD  n  GO  CO  t- 

§^  Ci  CO  GO  GN 
G^  CO  »O  CO  CO 


CiTfCiCOGO      CO  t—  <M  CO  T-( 
^CDt^CiO      (NCOOCOX 


O  —  GO' 

X  -t  Ci 
r-  CC  -T  CD 
^  "t  «*  -f  -t< 


CO  'M  CD  O  GO  CDCiCOCiCO  GO^CO 
CO  1^-  O  CO  »O  i-  X  Ci  Ci  Ci  X  r-  O 
CD  I-H  i-  'N  t-  cq  i-  T-I  i-  GN  i-  <?q  r- 


•^t  CO  CO  "^  T™H  CO  O  CN  Ci 

.   CM  CO  O  CD  t— 
CO  CO  t- 


COCiCOCO 
OCD^Nt- 


00  X  "^t1  Ci  Tt^      Ci  "^  Ci  ""3*  Ci 

CiO'NCCiO        CD  X  Ci  — '  -N 
-t  i.T  O  O  i~        O  »O  »O  CC  CO 


GO  O  GO  i-H  Ci  <M  O 

CO  i^  !"•  ^  Ci  kO  O 

CiThCi'*  GO  CO  t—  <N  t- 

•<t  1C  1—  GO  O  i-(COTj<COt- 

o  CD  CD  co  r- 


i-(  o  co  ^H  co 

X  »O  i—  t-  ^N 
CO  Ci  O 
T-I  <M  ^* 


^  CD  ip  GO  r- 

cc  x 

»0  ScO 

•*t   Tt   Tt    't   't 


X  CO  CC   X   X 

Ci  -f  Ci       O  O  iO  O 


i-  CDCOCir-l(M 

O  <O  »O       >O  iff  »O  CD  CO 

^«    T}1    -f 


X'tOO 
~fOCOi-H 

<M  -^  »re  t- 

•^  Tf  -t  Tf 


GO»O        CD5<1(MXX        CO 

3SS    ^^jii^    g 

;5J   I 


LOGARITHMS. 


<N   <N  i-H  rH  O  O   OS  OS  OS  00  00   I—  t—  SO  CO  CO   xO  >O  ^  Tfl 

^   ^  -*  •«*  T*  T*   CO  CO  CO  CO  CO   CO  CO  CO  CO  CO   CO  CO  CO  CO 


co  os  1-1  r—  os  r—  o  oo  <M  TH  10  co  1-1  co 

CO  OS  CO  iO  t-  OS  r-l  TH  <N  ;N  T-I  O  OS  t-  » 

GO(Mt—r-(»O  OS  -<*  GO  <M  CO  O  T*  1^  i-H  » 

OS  i-l  <M  Tj<  O  CO  GO  OS  i-H  <?}  TJ<  O  CO 

t-  GO  OO  00  OO  OOOOOOOiOi  OiOSOS 


i-l  i-l  »O  Tt<  O       i-HOOl-lO^t 

"I  OS  lO  ~H  r—      (NCOi-HiOOO 
jq  CO  O  CO     i-O^t-O 
t—  OS  O  i— i  CO 
O  O  i— i  i— i  i— i 

o  »o  »o  o  »o 


os  co  GO  10  i— 

l-H 
t- 

OS  1-1  <N 
00 

M< 


»o  os  t-  i— i  i— i  co  t— 

~  co  lr—  oo  or-  t—  CD 

<M  CO  O  "^  OO  !M 

00  OS  i-H  5<l  CO  >O 

OS  OS  OS  OS 


33 


t-  00  00  OO  GO 
•^  ^  ^  ^  Tj< 


-H   CX|  CO 

CO  GO  OS  O  C^        CO  O 

00  00  GO  OS  OS       OS  OS 

•^    Tf    Tfl    ^    Tfl 


OS  r-t  CO  i-H  O  Tj<  iO 

Ot— <M  GOCOt-T-i 

oo  TH  10  oo  <M  o 

O 


i-li-lCOCO(M  COOC^OSS^ 

T-HkOGO'-i't  COOOOSOSO 

t-i-liOO-*  OOtMCpQO 

^3333  ^^^^H^H 


I-H  oo  -H  GO  rj 

if5  t-  00  OS  in 

GO  OO  00  OO  OS 

•^          -^  -*   ^   T^    ^ 


^  CO  00  00       -TH  CO  "*  t-  CO      O  i-t  t—  OS  t- 

OSOOCO^        <7<IOSCO(MOO      TjHOSCOt-i-H 
<7<ICOO^        GOi-(»OOSC 

os  os  os  os      os  o  o  o  c 

•^  »O  >O  iO  O      »O  ut> 


OS  CO  CO 
t-  OS  O  <T4 


>O 


O  —  I 


t—  OS  t—  1-1  O 

>*^  u.^   xi'   -•»•  — i        00  O  ^  OS  lO 
t—  --  »!7  OS  CO       CO  O  -t  t—  I-H 

CN^iOCOOO        OS^H^NCOO 
OSOSOSOSOS       OSOOOO 


WtCO 


cocoeocow      eoeoMMM     eowcceo 


LOGARITHMS. 


CO  <M  <M  -H    rH  -H 


rH  -H  O  O  OS    OS  OS  GO  GO  GO    t— t-  CO  CO  CO   »C  »O  O  "f  Tj« 
CO  CO  CO  CO  <M   (N  <N  <N  <M  <N   <N  <M  <M  <N  S<1   <M  <>}  <M  <r*  'N 


O    rH    -*    'N   I  — 

rH  "*  CO   X  OS 


SrH  rH  rH  rH 
CO  CO  OS  <M 

rH    (M    CO    «tf    CO 

(M  'N  (M  (M  C<l 

O  O  »O  >O  O 


rH-^COOS         <M"tft-CSrH 

lOiOt-GO       OrnrNCOiC 

ss  s^^ss 


CO  O  CS  (M  i 
t-  X  OS  '-H  <7<l 
CM  (N  <M  CC  CO 

O  O  O  O  »O 


CO  r*  GO  t-  CO        10 

— I   TK    CO   GO    O          rH 


CO  t-  CO  »f5  ^ 
CO  ^  CO  f-  i 


OS  O  t—  I-H  <M 

(M  GO  G<l  1—  T-H 
GO  O  CO  O  X 
"  rH  (>1  CO  "* 


»0  ^  OS  O  t- 

^  TH 

GO  i-l 
(M 

co  co 

O  kO  O  10  O 


CO-HCOCOCO       COiOO(M 
O  CO  CO  CO  CO        kQ  ^F  CO  «H 


t-  O   Tf<   t— 

CO  »O  CO  t- 

O  id  O  iO  »O 


rH  O  >O  CO  ^  t-  CO 

CO  CO  CO  <M  rH 

(NOGOrHT^  t-O-- 

5<l(Mi>liT<l<N  <N(MC<I 


rH    ^  1—  OS  rH 


OS  lO  CC  t-  (M  CO  rH 
>O  -^t1  7^  O  GO  lO  "~ 
I-  O  CO  CO  GO  rH 


GO  GO  ^  CD  •*   GO  GO 
rH  -H/l  i^  OS  rH   (M  CO 


3 


JO  O  lO  »O  O 


<N  <N  <7<l  (5^   (M  <M  (M 


'8 


t-  (M  CO 
«O  GO  O 
>O  CO  t-   OS  O  rH  (M  ^ 

^^SSS   g^^ 


»O 


O  »C  O 


CO  W  CO  CO 


r-OOOSO      HWCO^JO      O**00540 

CO  CO  CO  CO   CO  CO  CO  CO  CO   CO  CO  CO  CO  CO 


100 


LOGARITHMS. 


OS  1—  <M  -rji  <M 

x  x  x  i—  CD 
ic  i~  cs  —  co 

X  OS  O  <M  C- 


X  O  OS  •<*  1— 
CO  O  X 

i—  ~  ,-**  ™   -f  iC  i—  OS 
O  rH  'N  CO 
,_  ~  CD  CO      fr-  t-  ^  J 


T-H  ic  x  o  <T<I     -t  CD  r—  GC  a 
T*<  co  x  —  '  co 


Ci'NrHt-O         OS005OIT- 

'M      ^St^OiT-i     co^ 

CO        -^f  O  ^  I--  OS        O-Hf 


(M  »C  I—  OS   rH  CO  CD  X  O 
t—  X  OS  O   'N  CO  ~f  iC  I— 


B   rH  CO  iC  CD  X   O  G 

O    XCSOrHCQ    T*K 


XOXXiC       XXiCXOS 


. 

rHCOTtlkCCO        1—  OS  O  rH   - 
»C  »C  iC       »C  iC  CO  CO  CO 
iO»C 


r- 1-  3  xxicoso 

•*  iC  CO   CO  CD  CD  iC  »C 

-  =  r  -*  '~ 


J  t-  OS  rH  5q 

t—  X    OS  O  rH  CO  -t1 
CO  CO   CD  t—  t—  t-  t- 

iC  »C   iC  iC  iC  iC  iC 


O'TlCCO         t-OSCOrH^ 
OSOrH'M         CN^N'N'NrH 


M5»OW5K5»O       lOWiCWO       Q505D 
OQCOOQCOCO        COCOCOCOCC        COCOCC 


LOGARITHMS. 


101 


iC  iC  iC  -*  ^       TfT^cOCOCO       "M  <M  <M  -N  TH       TH  -H  O  O  O       O 


O  CS  C5  CS  X 


s  s 


iC  iC  O 


CS   Tj<  I—  t-  -*   X   CS   X 

!M  CO  CO  CO  CO  <M  TH  O 

t^XCSO  ^(NCO-^^t 

co  Tf  »o  tr  x  cs  o  TH  ^ 


iC  iC 


o  oo  TJ<  t- 1-     T^OSO 

CS  CS  CS  CS   CS  CS  O  O 
iC  iC  iC  »C   >C  iC 


. . 1C  O  <M  CS 

CS  O  O  p   Q  Q  CS  X 

TH  O< 

cs  cs  cs  cs  cs   -  •" 

»C  iC  1C  »C  >C 


iC  CO  t-  X  CS 


O  iC  iC  iC  iC 


I 


CO  CO  CO  CO  CO   CO 


102 


LOGARITHMS. 


J-—  r—     i—  t-  co  co  co     cc  >c  >c  ic  >c     -ti  -*-*"*  co     co  co  co  -M  CM 

00   00000   000  00   00000   00000 


x  r-~*  x  cs 

Slili 

-'•v   *,~.  r~   rf-',  ~r* 


III?     1A»)     UJW 

iC  CO  t^  < 


co  co  co  co      co  co  cc  cc  co 


CM  7<1  O  CD  OS  O  GO  CO  t-  GO 

CO  t—  T-I  -*f  1—  TH  CO  CO  X  O 

O  O  TH  i— I  T— I  CMCMCMCMCO 

OTHCMCO^ti  iCCCt-XCS 

CM  CM  CM  CM  CM  CM  'M  'M  'M  O-l 
CO 


OCSCOi-H'-M        ri  r-  ^H  C<J  O 
CCCOCC^CO        COCOCC'COCO 


000 

-t  cs  cc 


*M  ?<l  CS  CO  iC 

O  GO  iC  CO  O 

iliil 


t-  -^  o  co  ,-M 

d  co  -*  ^t  ic 

OS  O  —  CM  CO 


3£ 


co  CD  cc  co     cococococo 


Tt<Ti<THCDX       t f^XOCS  iCCSTHCSCD 

CS  t-  iC  'M  OS  CO  CO  OS  C_;  TH  1~  >l  X  -M  1- 

t*—  X  CS  O  O  TH  CM  CM  CO  "^  "^  O  O  CO  CO 

CC  -t  iC  1—  X  CSOTiCMCO  -f  iC  CO  1—  X 


cocococo      CDCOCOCCO 


THOOrM-tCO  0-t<COiCTH 

CO  (M  OS  iC  i-t  1—  T-J  1—  C^  1  - 

~    TH  TH  ^1  CO  CO**-^iCiC 

T-H^NCO  ^iCCOt-00 


t=  i-  r-  x  x 

S§7i?l^ 

CD  CC  CC  CC  CO 


iCCCiC^NCO 
TH  iC  CS  CO  CO 
cOCCCOt-t- 
OSOrH-MCO 


GOt-COCC 
CS  ^N  iC  r- 
t-CCXCOCS 

OO 


CO  ^  Tt1  -^t 

CSOTHCMCO        TtiOCC^- 
THCMCMCM'M        CMCMCM 


CM  CM 

X     TH 

CM  CO  >.•»  •-•- 

co  r- 

CM  CM  CM  CM    .  . 

cc  co  'CO  cc  cc 


LOGARITHMS. 


103 


r-<  i-l  i-l        rH  O  O  O  O         O  Ci 
OOO       O  O  O  O  O       OOi 


X  CO  CO 
CO  CO  fM  !M  <M 

O  CO  I—  X  Cl 


O  — '  O  CO 

T-H    X   »O    r-H 


oo  oo  Ci  a  x 

^  »M  -M  'N  -N 

O  O  t—  X  Ci 


i-H  (M 


CO  Cl  CO  »O  »O       'Mt-CSOX       CO  t~  X  CO  CO       t-  O  O5  t 

(MCOOCOt-       XXXOiX       X  t—  CO  O  ^       (M^nXC 


i-(  C<1  CO  •**       »O  CO  I—  X  C5 

co  co  co  co     co  co  co  co  co 


g* 

qHOlCO^         ^^^^^ 

ocococo      cococococo 


CD  CO  CO  CO  CO 


._  OO  F" 

^o^oic 

O  CO  CO  CO  CO 


5  rH  CO  CO  -H 

:  x  i-  co 

5  O  iO  O 
OS  O  1-1  -M  'I 

a3333 


X  CO  CO 

s^£ 

— .  v^^  rH  C^  CO 


104 


LOGARITHMS. 


tOiOtOiO-^       ^t-t-t-^ 
OSOSOSOSOS       OSOSOSOS 


(M  £-    _  _ 

OS  CO   CO  t-  1—  «D  CO 
t-  CO   OS  O  rH  <M  CO 


to  ^t<  ^  co  CN 
•*T  to  ?6  Jr-  op 

o  co  CD  CD  o 
o  co  CD  co  co 


tO  OS   O  OS  t-  CN  CO 
<M   tO  CD  CO  O  ^ 

-   »  t-  co  co  to 


-t  CO 
<M  tO 

(Mi-iOOOS  CO  1-  CD  CO 

§O  i-i  ^  <M  co  -*  to  CD 

i—  r-  t-  t-  i—  t-  i—  i-  i— 

CD  CD  CD  CO  CD  CD  CD  CO  CD 


^ti  CO  CO  !T<1  — ( 
-t  »O  O  t-  CO 


SOOL'OSCO  !>•  CD  Jo  >*0  ?? 

OS  O  O  ^  <N  CO-ftOCCt-* 

co  i-  i—  t-  i—  r-  i—  i-  i--  t- 

co  CD  co  co  co 


OOSiOOS  T— i  —n  OS  »O  CO 

— I  CO  <M  i>-  CO  CO  <M  1"-  — i 

COt-t-CO  CDtOiOTfT^ 

tocct—  co  os  o  i— i  cq  co 

>O  tO  iO  tO  tO  CO  CO  CO 


o  ct  r-  co  co 
co  ct  co  t-  o 

CO  7<l  ^  i-H  — I 
^  >O  CO  t-  CO 

CO  CO  CO  CO  CD 

co  co  co  co 


CO  CO  CO  OS  CO 

cc  o  -t- 1—  T-H 

(N  7<1  r- 1  O  O 

tO  CO'  I-  CO 


fN  --M  O  CD  r-> 
CO  O  CM  CO  iO 
CO  'M  T-I 
CO  -^  tO  CO  t^ 
l^  t-  t-  1—  i— 


tO  »O  tO  »O 
CO  CO  CO  CO 


CpCOiO-HtO  COCOCD(>JCO 

os  -Tf  os  co  i^~  i—*  to  os  o^ 

cococ^o  CDCDCDCOCO 

CO  CO  CO  CD  CO  CD  CO  CO  CO 


osr~cor~  oocotoos 

OSOT-ICN  CO^tOCOt^ 

cot^t^t^  i-  IT-  t-  r-  t- 

COCOCDCD  COCOCOCOCO 


O  rH  O 
to  O  tO 


CO  -^  <N 

CO  C^l  CD 

9  S  S  s  8 1 1 

co  «  co  co   co  S  cc 


•  CO  tO  CO  ^t1  I-H  iO 

co  co  os  <M  to  r- 

i—  co  to  10  TT  co 

1-  CO  OS  O  T-H  <M 


tOCDiO'Nl—       CSOOliOO       (NCO-^HCO 
to  O  iO  O  -^       CO  CO  'CD  O  "^       i—  O  CO  to 


OS   -H   O  -        -  . - 

COCOOS^O  tOOOO 

COCOtMCN'N  i-HrHOOOS  COCOl-l-CD  -_  -_      .---. 

-^  iO  CO  i—  CO  OS  O  — <  <M  <M  COTOCD1—  COOSOr-(<M       COrfH 

»O  tO  >O  tO  iO  iO  CO  CO  CD  CD  CD  CD  CO  CD  CD  CO  CD  t—  t—  I—   t-  r-  t-  t-  t"- 

cococococo  cococococo  CCCDOCDCD  cococococo      COCOCOCDCD 


COtO^'Nt—       O-HO^-(N       iOCOiOC<l 
t2  CO  OS  to  O       CO  ^  CD  O  ,  O       OS  CO  I-  ^ 


t '  co  -co 


CM  C^l  -H  r-l  r-l 

•^  iO      " 

tO  iO  tO  tO 

CO  CO  CD 


cocococo      cocococo 


il 


»^  ao 

S3 


LOGARITHMS. 


105 


Oi  —  •   -•  Gi  —  - 


»ooi~r~ 

T-H  O  Ci  00 
GO  Oi  O  O  —I        (M  CO  -t4  O  O 

t~r~xccoo     o>xxxoo 
ooooo     ooooo 


r-  r-  o  co  GO     T-H  co  s<i  o  o 

IT*  .C"*  IT*  t'*  to       to  iO  "^  CO  TH 

r-  tD  »O  rj<  CO       <M  i-i  O  OS  00 


ooooo 


lib  >O  1"*  t*  GO       GO  GO  GO  GO  t^ 

co  Ci  Si  o  — t     <?q  co  -t  3  « 


t-OiO-^(M 

i-iOCiGOi- 


COOiOOr-l         i— i 


F-  co  co 

ooooo 


—  1—  O  »O  rf 
t-  CO  Oi  O  1-1 
1^-  1^-  t^»  CO  00 

O  O  O  O  O 


^NS<|(M(M(M       1-1  O  Oi 

CO^i-lOOi       COt— :~ 
C3  CO  Tfi  >O  ifi 


§Oi    Oi     O     C^l 
O  S  "t  CO 
t-  CO  Oi  O  ri 


§00  CO  O  *-H 
-H  (N  CO  -fi 
10  rf*  co  ;N 

I-  X  Ci  O  i-l 


iO  t-  I—  »O  <N  O  Oi  O 

•^-l-^-t-*  CO'N'N 

TI  ©  OS  CO  t-  OiO'* 

(MCOCO'tiO  Oi—  CO 

•  co  co  co    " 


O  Oi  TH  O 


- 

O         rH   i-l 


io6 


LOGARITHMS. 


t-  co  CO  CO  CO   CO  CO  »O  »O  »O 


t-  <N  Tj<  »O  iO  <N  GO  CM  iO  CO  iO  CO  OS 

TH  GO  M<  O  CO  GS|  t-  CO  GO  CO  GO 

CO  ~H^  CO  C^  O  OS  ^*  CO  "^  CO  TH 
- 


rHGMCO^        •»*  lO  CO  IT—  X 
0000       OOOOO 


TH  »O  X  OS  OS       t-  CO  I— 


8TH        OGO^fOSiH        CO5q 
lO        O  "*•  OS  CO  GO        CM  CO 


«D^*         CO  TH 

T-H  (M       CO  -* 


-*  CO  »O  ^   TH 

O  CC  CO  Oi  7<l 


rfi  OS  <M  CO  CO 

-*  O  t~-  CO  OS 
-t  CO  TH  O  GO 


=22 


00  GO  CO 

si  sss 


TH   CO   CO   -H   ^ 

GNMO  GC1  T^  « 
rj<  CM 
1-  X 


X  S<l  CO  t—  t— 

»o  ^  x  -^  o 

CO  <7<l  O  OS  X 
i^  ,_!  jq  7<j  ^O 
0000 


TH  TH  OS  lO  O  CO 

COOCO'NO  OSt-cb-t^CO       THOiX 

-<^  iO  CO  £-  X  XOSOi-HCM       COCO'* 

p  p  p  o o  op 


% 


X  >.C 

GN  10 

CO  TH  O  X  CO 
t-  X  OS  OS  O 

1—  - 


-^-T 

oooo 


OS  CO  TH  -H/l  CO 

t—  CO  OS  "*  OS  ^fOi-fX'M 

OOOOO  O  O  TH  TH  TH 

t—  t—  t—  t—  I—  t—  1^-  I—  Jt—  t» 


OOOSCDi-H 
t— r-lTtGO'N 


oTo 


^ 


E 


OS  OS  GO  iO  O  ^  O 

GO"^  OO5<lt~-(N  i~-  'N  1—  — tO 

COCDJO  •<*  (M  TH  OS  CO  OOCOC^O 

t-  i—  i—  IT-  r-  IT-  t—  t- 1—  t—  r-  i— 


§O   TH    O 
r^  iQ  GO 

t-  r-  x  os 


THt-THCOCO       CMOSiOXO       THOt~<NCO       XOsr- 
TH  I—  Tt*  O  CO       CMi-~COX^       OSTfiXCOt-       THIOCS 


t—  CO  Tfl         CO  TH 


OS  CN 


TH  CM  CO 
000 


t- 1-  i—  r-  t- 


CO  iO  CO  »O  (N  X  (N 

^OiO  OOO'UCSS 

OSXCO  lOOIfMOX       t-O^CN 

38ooo  §§°^     ^co-r.o 


IO    Tf    TH    t-  TH-^ld^CSI 

t-  TH  »O  CO  (MlOOOTHTt< 
OS  IT—  iO  ^  GM 
CO  t-  GO  CS 


OOCOOS         XOOCOTHt-         <M  1—  rH  CO  O 

OrHGNCO        "^flOlOCOt-        GOOSOOrH 
OOOO         OOOOO         OOHrH 


TH  (M  TH  'OS 
t-  O  CO  iO 
CO  O  CO  TH 
'  t-  00  OS  O 


OI^OOCSO        H 


LOGARITHMS. 


107 


_ <N   71   — '   — I   rH          1-HrHrHrHO          OOOOOS          OSOSOSO5 

X  X  X  X  GO      XXXXX       XXXXX       XXXXt-       • 


CO  t-    OS  O  OS  t-  CO    X  rH  CO  ^  CO 

rH  rH    rH  <M  ,-H  rH  rH    O  O  OS  X  1— 

§Ss§§ 

CO  CO  CO  ^  "*T 
t-Jt-  t-  t-  t- 


_ 
t-  iO  CO  i-l 


O  CO 

:?  :£ 


Ci— 'CO       CO  i-i 


CO  t—  OS  O  OS   t—  CO  t—  O  71   71  O  t—  71 

7^  O1  75  71  7?   71  71  7-1  7-1  71   71  CO  CO  CO  CO   CO 


OS  O  OS  t- 

»o  :c  o  uc 

1—  »O  CO  I-H 


^H  ri  co  co  't 

n  'N  'M  n  (N 


Oi-HO  XOOX»O  O  ^f  t-  XI— 

t-       JO  CO  rH  OS  t-  if5  -t  77  OS  b-  »f?  CO  rH  cs  1^ 

X        OSOrHrHC^  CO~t<»O)!7O  1—  X  OS  OS  O 

C<l        (MCOCOCOCO  COCOCOCOCO  COCOCOCO-t 


>C  — t  10       X  O  O  X  10       rH  »0 

.•*    m   *~>  M.-         w*<  w*  O  O6  1— I          rH    7^    71    rH   rH    .      1—t   O 

TJpXO       "^  71  —  OS  t—       lOCOrHOSt-       >OCO 
71COCOCOCO       COCOCOCOCO       COCO 

t-  t-  t—  t- 1- 


CO  O  X  O  (M 
71  O  X  t-  - 

rH  71  71  CO 

71  71  71  71  <M 


X  uO  O  -*  ^5   t-  O  -t  O  i. 

Ot-OSQrH    ««3f|2; 

CS  O  O  rH  74 
71  CO  CO  CO  CO 


O       (N  »  OS  -H  rH 
>O        O  -t*  CO  CO  <M 

COrHOSt-tC 


X  OS 
t-  t- 
<M 


t-  t-  t—  t-  t-  t-  t-  t-  i—  t—  t-  r-  r- 1- 1- 


<M  t—  O  rH   C<I 


m 


rH   TjH   O   «O 

I-H  CO  O  t- 

X  O  -f  Ol 


§rH    71   ^   >O  ^O    t"—   X   X   OS  OS 

OS  t—  O  CO  ^H  OS  r—  lO  CO  rH 

O  O  t-  X  OSOSO— 171  COCO'HHiO 

(N71717171  7171COCOCO  CO 
t-  t-  t-  t-  t- 


OS  t—        CO  ^  rH  (M  CO 

Os  OS      os  X  X  I—  :c 

rH  OS  t— 
t-  t-  X 

COCiOCOCO       COCO 


OS  t—  WO  CO 
t-  X  OS  O 
CO  CO  CO  ^ 


CD^*QOCSO      — •   **    ^   -*   '*!        O 
NStNNM      MMMMM         M 


io8 


LOGARITHMS. 


OS  05  GO  00  GO 


^1-COCDCO   CO  CO  CO  CO  tO 


OS  CD  <3<l  t— 
~  CO  Oi  "<f 
GO  tO  CO 

CO  T(<  »o  iO  CO  t-  GO  GO  Oi  O 
tO  iO  tO  tO  tO  *O  tO  tO  tO  CO 
t—  £—  t—  i~  t— 


c^ 


os  co  co 

r-i  iO  CO  ~H  OS   CO  •<*  <M 

TH   CM  CO  Tt<  Tt<         tO  CO  t-      ; 

t-  t-  £-  t~  t-      i-  i—  t-  t—  t 


^M     ^  ^    CO    TH    t-    <N 

CM  GO        ^  O  O  TH  t- 
<M  O  t—  tO  ^N 

t-  GO  GO  OS  O 
tO  tO  iO  iO  ~~ 


OStOGOO  THi-HOStOTH  Tfl  t-  GO  GO  CO 
GO  i—  iO  ^  <N  O  t~  tO  CO  O  t-  -*  TH  X 
"*  CM  Q .GO  CO  -*  -H  Oi  t—  iO  3<J  O  GO  tO  <ra  TH  CA, 

»O  »O  lO  iO        iO  »O  iO 


to  CD  i—  r-  GO 

T}*   Tf<    Th   ^    TfH 


GO  1^*  tO  TH  CD 

CO  (M  GO  -^  Oi 

TH   OS   CO   T^   TH 

t-  t-  X  OS  O 

tO  tO  to  iO  CO 


CM  T-H  OS  GO  CO       T^I  7^  O  I'*  tO       ^  O  f"*  "^  O  lr^  -^  O  t^*  CO  OS  tO 

COTtlT-HOSt^       tOCOTHGOCO        "^(MOSJ^-tO  (NO  'X  iO  CO  OGO 

THCNCOCO^       tOCOt^J—GO       OSOOi-HCM  COTf-HiOCO  t-  t- 

-«t  »O  >O  tO  iO  tO  tO  tO  tO  iO  iO  »O  iO  tO 


tO  iO  CO  O  CO  O  CO  "f  "^H  OO 

CO-t^cMO^-  tOCMOSCOCO 

>OCOTHO1CO        Tt<CMOGOiO  COTHXCO^ 

TH-MCOCO^ti       tOCDt—  I~GO  OSOO-H'N 

T^    "^    T^H    Tj<    T^  Tf^    T^    Tfl    TJH    TfH  -^    ^Q    jQ    lf$    tO 


CDO^CO       COtOCOO 
COCOOStO       THl^-COOS 

-  i-  ^  ^      -  -      


S3  23  8 


T   CM  <—  '   C7j 
TH  <?<l  CO  CO 


t-  i~  t-  r-  L- 


GO  »O  CO       THGOCOTfHTH        Oil—  rflS^Oi 
OTH-N        COCO-^tOCO       CDt-GOOiOi 
iO  iO  iO  tO  O       tO  iO  tO  tO  tO 


CO   -H  Oi  t—  tO 
TH  71  7^1  CO  ^t 

I—  I—  t- 


Oi  OS  GO  lO  TH 
O  GO  CO 
CO  O  GO  CO 
iO  CO  CO  t-  GO 


tO  GO  O  O  OS 
OS  CO  Tf  TH  t- 

TH  OS  1^  l 


•^  TJ<  (M  OS  ^ 

CO  OI  GO  CO  OS 

GO  O  CO  TH  GO 

co  r—  GO  os  o: 


tCO'Nt^OS  THT-HTJt-CO  GO-HCOCO'M 

OSXCC'*  COi-HOSCO-^l  TH~ 

OXCOTf  cMOI^-tOCO  -H 

TH  n  CM  co  -t  10  co  co  t-  GO  os 

-t-t^-f'Tjl  ^^-fl^^  T^-t^ 

t-  i—  t-  i—  t-  t-  t- 1- 1-  r- 


O  TH  <7<l 
iO  tO  iO 


O  CO  — i  tO  t-  GOGOCDCOOS 

t-COOCDCM  GO"*  — 

OS  I—  O  CM  O  t--  lO 

n  co  -t  to  CD  co  i— 

tO  >O  tO  tO  tO  tO  tO  iO  tO  tO 


GO 


T-H        CO  CO  ^M  Oi  ^O 


3 


tOOSTHCMCMOGOCO 
COi-HOGOi-       tOCOTHGOCO       •^tii-HXtO'N       OSCD.T^XtO       T-H  t-  *0  GO  Tfi 

^NOXtOCO       THOSI t^(M       OXtOCOrH       XCO-fr-iOS       J^TfifMOil^ 

TH  '2^  'M  CO  -t       tO  >O  CO  I—  GO       OS  Ci  O  — '  CM       <M  CO  -H  >O  iO       CO  *-  X  X  OS 

10  >0  >0  iO  >0       »0  tO  iO  »0  to 


(N  OS  >O  O  CO 

O  CO  CM  TH  OS 

TH  OS  L—  >O  CM       OX 

TH  TH  <M  CO  -^       iO  iO 

^*  TfH 


tOtOTMcMX       COCDXOSX  CO 

THGO       COCOOt-TJ<  TH 

t^  X       GO  OS  O  TH  ^  ^| 

TjH  -HH       -^  r^  O  iO  iO  lO 

t- 1-  t-  r-  t-  t-  t- 


O  CO  GO  -M  tO 


3  3 


Oi   O       H  N   «  ^   10 
O  t'      r-  r«  r-  r-  r- 

10  10       10  10  10  10  10 


LOGARITHMS. 


109 


CO  CD  «C 

r-  t-  r- 1-  r-     t~  t- 


rH  rH  <N  CO 

CD  CD  CO  CO 


lO  O        lO  GO  O  rH  O        OS  CO  C^  CO  O 
>  i^l^        GO  OS  OS  O  rH         C^^CO^flO        O  CD  t-  GO  GO 


»O  IT-  GO  CO  i-       rfl  O  iO  OS  rH        <N  <M  O  i—  CO 
t**  C<1  £"•  G^  IT*        C^l  t~*  rH  lO  O        "^  GO  ^1  lO  OS 


O  CO  rH  >O 
«O  OS  -*  GO 


QOGOCD"^O        O  GO  rH  <M 
^GO<M       COO^GO(M        lOGOC<J»O 

rti»'oCDCDt—       GOOSOSOrH        rH<MCO"<* 

cococococo     cococot-t- 
t-     t-  t-  t-  i- 1- 


3£§££ 

O  CO  t- t-  QO 


"*  t—  GO  GO  t—       1C  rH 
C<l  t-  <M  t^  CM       t-  G<1 

O  rH  C^l  (^1  CO        "^  lO  >O 


rHOGOOSOS  GOlO(Nt-rH 

_  _  GO    r—    ^*    |^—    O  CO    CO    OS    rH    ^ 

t-^        rHOSCO^r-t        GOCOCOOGO  iO<NOSr--<* 

CDt-        QOGOOSOrH         rHG<ICO'*^  >OCOCDt-QO 

co  co     cococot-r~     t-t-t-t^. 
t-  t-  i—  t- 1- 


§S 


<M-^COCO  »OCOOS-^G< 

OSCOI—       O^l^OCO  CCOSrH-^CC 

rHQOlCCOO        GOOCNOt-  ^rHOSCDa 

GOGOOlOrH         rH(MCOTt*Ttl  lOCDCDt~G< 

cococot-t-     t— t-i—t-t-  t-t-r-t-t- 


^   £££££ 


S 


rH  GO  CO  1^  O 
lO  OS  "^  GO  CO 

c<i  OS  t- 


t-  rH  iffl  GO 
OS   I—  Tf    rH 


GO  O  O  OS 

iM  CO  OS  rH 

CO-HGCiO  COOb-^G<l 

'TqcOCOTjH  OCDCOt-GO 


<?}  CD  GO  OS  GO 


CO  CO  O  CO  CO 
t-  -M  CO  rH  lO 

rH  OS  CD  Tf  rH 

-*  jt  o  co  r- 
1-.  t-  i—  t-  t- 


>Od        t-(M»OCOb-        CD-^rHI-rH 
GOC<|lOGOrH       -^1  t—  O  0^1  >O 
(M  OS  1-  ^  rH 


TOrHQO        »OCOOt->O         (M  OS  1—  Tf  rH 

OSOO        rH(MCOCO-^        »O  »O  CO  t—  GO 


£££ 


r-  oc  05  o      H  « 
r-  r-  r-  00      QO  00 

tQ    tQ    IQ    >C         10    >4 


O  O 
1Q    IQ 


no 


LOGARITHMS. 


<M  <N  <N  (N  CM 
JL1—  j 


O  O  O  OOS 


»O  CO  O  CM  CM 
T-I  TH  CM  CM  CM 

CM  OS  O  CO  O 
O  TH  CM  CO 

—  -  — .  — -  OS 

t-  t-  I—  t-  t— 


CO  "<*  iO  »O  CD 

OS  OS  OS  OS  OS 

1—  Jt-  t— 


^  GO  O  C^  CM 
TH  GO  »O  CM  OS 
OS  OS  OS  OS  OS 

Jt-  r-  i-  Jt-  Jt- 


O  CO  O  t- 

co  •*&  >o  10 

OS  OS  OS  OS 


CO  GO 

GO 

CO  ^  »C  iO 
GO  CO  GO  00  GO 


GO  GO  O  CO  OS 
CO  Tj<  »O  O  CO 

»o  IM  os  co  co 

CO  t—  i—  CO  OS 

00  CO  GO  CO  CO 


-#  GO  O  TH  CM 
t-  I—  GO  GO  GO 
O  Jt—  "HH  -rH  GO 
O  O  TH  2<)  CM 

os  os  os  os  os 


o  CM  cs  to  co 

CO  HH  H<  >o  CD 

os  os  os  os  os 
t-  Jt-  1—  t- 1— 


GO  OS  OS  CO  CO 
O  CM  -H/i  CO  GO 
CO  O  1—  ^  ^-H 
OS  O  O  TH  CM 
t-  CO  00  CO  CO 


(N  GO  CM  lO  t- 

•n  co  Tt  10  »o 

GO  GO  CO  CO  GO 


t-  i—  »o  co  os 

co  t-  .x  os  os 

-*  i-H  CO  iO  (M 
CO  t-  t- 

GO  GO  GO 

i^  r-  i- 


"*  t-  O 

~  O  T— I 


co  4t  ^t  >o  cb 
os  os  os  os  - 


lO  Jt—  OS  TH 


i-i  CD  i— l  Tf 
CO  ^  CO  t— 
GO  lO  'M  OS 

fN  CO  rfH  -t 
00  GO  GO  00 


OS  O  TH  CM  CM 
CO  -H  CO  lO  CM 

CO  t-  Jt-  CO  ~ 
GO  00  00  GO 


.   .  -t  CO  CO  CO  'N 

COCOOt-  HH  -H  CO  O  (M 

O-r-triCM  CO-*-t»ccO 

OS  OS  OS  OS  OS  OS  OS  OS  OS 


•HH  CM   OS  IO  OS 


^H  t-  co  r—  os  —H 


-^  O  1~  CO  GO 

r-  ^-  co  co  10 


t-  £-  1-  t-  1- 


-H  CO  CO  CO  *H 

OS  ^H  CO  lO  t- 

QO  O  iM  OS 


GO  CO  CO  TH  CO 
CO  O  TH  CO  "^ 
CO  Tt<  rH  CO  >O 

CM  CO  TJH  — tn  »O 

GO  OO  OO  GO  00 

t- 1- 1-  Jt- 1- 


-  t- 1-  i—  t- 


TH  O  t-  ^  OS 

CO  CO  CM  CM  — ( 

CM  OS  CO  CO  O 
CO  CO  ^f  lO  CO 

OS  OS  OS  OS  OS 


;M  o  GO  ^  os 

CO  CD  iC  »C  ^t4 
TH  GO  ift  <M  OS 
CO  CO  "*  >C  »<? 

OS  OS  OS  OS  OS 

t-  t—  t- 1-  t- 


co  os  os  os  os 


CO  CO  ^t  »O  iO 
OS  OS  OS  OS  OS 

1-  I-  I— 


O        H    N    CO   Tlj 

3     o  o  3  » 


LOGARITHMS. 


Ill 


;_.  _.  _ .   Oi  --  --  *  GO 
CO  CO  CO   CO  CO  CO  CO  CO 


ggggg    gg££S 


'£ 


T-I  CO  CO  71  O  t-  CO  X  <N  »O 
i—  >O  CO  1-1  OS  CO  •<*  1-1  OS  CO 
O  t—  -rf  ^H  t—  ^  i— I  X  "<f  i-H  X  lO  i— I  X 


O  -t  70  71   i-H  OS  O>  CC  JO 
I—  "t  I-H  CO   >O  i-H  CO  O  vTJ 


t-  1—  CO  OS  OS 


I-H  I-H  71  CO 


OSOSOiOSOS       OOOOO 
~  t—       CO  CO  CO  GO  CO 


»O  CO  t-  CO  iO 

co  i~i  os  i—  >o 

OS  CO  71  OS  CO 

CO  -t  »O  »O  CO 

X  X  X  X  X 


Tf  X  TH  CO  "tf  "*  CO  i-H 

CO  O  X  O  CO  O  i~  ^t<  I-H  X 

CO  O  CO  CO  O  i—  CO  ~ 
t^~  X  X  OS  O 

X  X  X  X  X 


S££ 


X  X  X  X  X 


IT*  X  OS  X  1^*      "^  I~H  CO  O  *^  CO  Jt^*  i^*  CO  "^ 

-fTlOX       CO'ti-HOSCO  COOt-^^H 

iO7lOS»O       (MO5CO71CS  COCOOSCOCO 

OCO       t-t-XOSOS  ®^^^^ 

XX  X  X  X  X  X 


XOi-ii-HOS  t-COOSCOCO 

OSXCO-*i-l  OSt~--t71OS  _     ,    .   ._ 

t-  rf  1-t  X  ^O  i-H  X  »C  71  X  lO  71  OS  O  71 

CO  -t  »C  i-O  CO  t-  1—  X  OS  OS  O  T-I  T-  <N  W 


CO  GO  CO  CO   GO  GO  GO  CO  CO 


iO  CO  I—  CO 
T*<  CO  71  — I 

O  71  OS  CO 


73  lO  t—  GO  t— 

CO  t-  CO  >C 
t-  -t1  1-1  X 

CO  1—  GO  X 
OS  OS  OS  OS  OS 


©T^COCOi-l 
CO  --I  OS  1—  »O 
i—  -^  O  t  --  f 

i—  i  71  CO        CO  -t1  »O  O 

GOCOCO        COCOCCGO 


'TlOOCOt- 
OS  1—  CO  ^ 

OS  CO  CO  O 

O  i—  i  71  CO   CO 


CO  CO  GO  CO  X 


CO  CO  OS  iO  OS       (M 

CO71OOS1—       CD 


ooo     ooooo     oo 

XXX       XXXXX       XX 


O  i— I  -H  C<I  CO 
GOGCGOGCGO 


CO  CO  GO  OS  i 

CO  t-  X  X  < 

OS  OS  OS  OS  i 


Jt—  -*   i-H   CO   O 

CO  iO  ^  71  1-1       

I-H  X  O  71  OS       K7  >1 


O  CO  I-H  -^  t—  OS  OS  OS  t- 

1^*  "^  71  OS  CO  CO  O  t~*  "^ 

CO  CO  O  CO  CO  ^^  IT*  CO  C5 

*  ^  OS  O  I-H  T-I  71  CO 

X  X  X  X  X  X 


"*  t-  OS  O  OS 

;e  co  7i  71  o 


X  X  X  X  X 


1  ^H  O  X  1-  10  CO  71  O 

51—  TjH   O  t—  -t   l-H    CO    O   • 

5O-^(M7-1  CO-t^tO 

ooo  o  o  o  o 


CO  — i  X  CO  COOt—^i-l 
'  >a  71  X  ».O  iTl  OS  »O  71  OS 
I-  CO  X  OS  O  O  r-n  71  <N 


ill 


112 


LOGARITHMS. 


CO  CO  »O  »O  >O       »O>OiO»O»O       KO  >O  •'t1  ^fl  "rt 


O  O  X  CO       <M  X  CO 

co  03  r-  co     os  -^  o 

OS   O          O    r^    ?5 


rH  i-H  O  X         1O  rH  »O 
rH  CO  -H   »0         O  O  OS 


^  ^  ^      ^-t^>ococo     t—  r-  co  os  os 

XXXXX       XXXXX       XXXXX       XXXXX       XXXXX 


O  CO  rH  1C  X 
-*  X  CO  t-  rH 
<M  X  »!7  rH  X 

r-  i—  x  os  os 

rH  rH  i— I  rH  rH        rH  —H  -H  ~H  GNI        fN  ?<1  <>J  C^l  <N        C<IO4'2<I'J^(N        <M  C<1  <7<1  ?>1  ^ 

XXXXX   XXXXX   XXXXX   XXXXX   XXXXX 


10  OS  rH 
,_,  f^.  TJI  ^,  t^ 

^  -^  10  co  co 


X 


-to     r-  co  o  2  co 

os  o     o  rH  ri  CM  co 


rH   1—   C<J   CO   OS 

CO  —  1—  (M  t>-  COXCOXCN  t-(T<JCO'-liO 

CO  CO  OS  CO  <N  OSlO^XlO  rHX"^rHt- 

o  — i  — i  o<  co  co-^tiiooo  i— r— xosos 

rH  rH  rH  i— I  rH         rHrHr-irHCq         (M^'M'MfM  G<I(M'M'^(N  (N^<J(M(M(N 

xxxxx  xxxxx  xxxxx  xxxxx  xxxxx 


-^co<MO 

r-^r — ho  - 

.t"**  CO  ^^  1^*  CO  O  CO  CO 

-*iOcoco  r-cococs 


COCOOi 
^»O»O 


rH  t-  <M  CO  OS        rH^(MrHOS  OrHCOOCO  lOCOCOlO'*  rHl— 

OCO'NXCO  OS  O  O  CO  TH  COrHCOrHCO  rH»O 

COOSCO(MOS  lO'TIOSlC'N  XOrHX'*  rHt- 

.t-t-XOSOS  Or-rHC<!CO  CO-tiiCiOCO  t—  t- 

rHrHrHr^rH         rHT— IrHrHrH  GN1'MC^'T<IC<I  'MC<I'>1<M<N 

xxxxx  xxxxx  xxxxx  xxxxx 


^rHCOOCO        lOCOCO»OCO 


xxxxx     xxxxx 


CO  rH  O  X 
X   ^   OS   Th 

SrH^S 

(N  <M  G^  (N 

X  X  X  X 


OSOOOSt-         rfO»OOCO         lOCOCOCO^         rHXCOXrH 

COCOOS-HiO       CO  (N  t—  CO  — 

rHX-fr-lX         ^  rH  t—  ^t1 


X  CM  t-  rH 

OS  CO  C^  OS 

OrHi—-NCO       CO^»OOCO       CD^-XXOS 

r^  T— i   T— i   ^^  i— i         T — i   T— i   -^^  r— i  ' — i         C^   ^-1   7^   ^1   '7^1         ^1   Ol   'M   C^l   C^l         (7^1   O^l   C^-l    "T^J  (7^ 

xxxxx  xxxxx  xxxxx  xxxxx  xxxxx 


OS  OS 


XXXXX       XGOXXX 


OS  »O  O  -f  GO 

OS   O   ^H   CO  rH 
CO  O  t-  CO 

O   rH    rH    (M 

!M  (N  <N  fN 

X  X  X  X  X' 


rH  rH  rH  OS 


t^  co  os  co  t- 

rH  CO  ^^  >O  OS 
OS  lO  C^l   X  "^t 

CO  t—  X  X  OS 
C<1  'N  T<1  C<l  'N 

XXXXX 


xxxxx  xxxxx  xxxxx 


rHXCOXrH 


X  X  X  X  X       X  X  X  CO  X 


CO  CM   t-  'N  t- 


•*  O 

s§ 


O     rH  N  «  ^  «5 

y.   r»  ^.  t.  r.  r>. 


LOGARITHMS. 


CS  <M  -*       r-  OS 
CD  7<1  OS  1C        rH  i- 
O  rH  ,—  cq 
-t  -t  -t  -h 

X  00  00  QO 


—i  X  -t  O 

"^  ->         - 


O   rH   rH    7 
X  X   X   X 


XXXXX       XXXX 


7-1  X  "*         rH  t-  CO  OS  1C 

rH   i— i  Ift         COCO^^lC 

XXXXX 


t—  X  t—  CO 

o  co  co  os 

rH    rH    (M    7-1 

CO  CO  CO  CO 
XXXX 


co  -t 

co  co  co  co  co 

X  X  X  X  X 


r-  cs  o  o  cs 

CO  OS  CO  CO  X 

o  £  i  1  2J 

CO  CO  CO  CO  CO 
X  X  X  'X  X 


X  >C  <M  1—  CN       CO  CS  rH  <N  7^ 

rH  -t  I—  CS  <M  Tfl  CO  CS  rH  CO 
O  CO  7-1  OS  »C 
CO  CO  -f  -rt  1C 


OS   O    l-H   t-   Tf 


CO  CO       X  O  7-1  iC  1^* 

rH    — I    CS1  T-1CO^-*»C 

X  X  X       X  X*  OO  OO  X* 


1~  X  X  l~ 

rH   »^    —fc    ^^ 

i"  co  ^-  — 

§  co  co  co 
XXX 


1C  7-1  X  CO  t— 

i—  -H   -t   X   rH 
71  CS   1C  rH  X 

co  co  -t  ic  ic 

CO  CO  CO  CO  CO 
XXXXX 


co  cs 

CO  1~  I—  X  X 

co  co  co  co  co 

XXXXX 


-t  rN  cs  »o  o  ^  i-  o  rH  <M 

o  co  <o  x  -H  co  o  x  o  <M 

co  -N  x  -*  -H  i^  co  os  co  (M 

OS  O  O  rn 


co 

X  X  X  X  00 


LOGARITHMS. 


§00000 
0000  tt> 


0  ^t  i-l  CO  CO 


1—  Cl 

Ci    O 

rH  OO  -tO 

rH  rH  CM  CO 

m  m  m  m 

xx  xx 


ci  -t  x  CM  in 


CO  CM  Ci 
CM  00  "* 


-. 

00  CO  00  00 


^.ninm     m  in  ,n  $  o     !§  £  £  £  £     $  S  §  §  § 

X  X  OO  X  X       XXXXX       XX  XXX       XXXXX 


rH  O 

in  r~ 

rH  t-  CO 


00  00  00  00 


r-  co  x  CM  in  i-  x  ci  x  t- 

co  -^t  ~t  in  in  o  o  o  o  in 

co  ci  in>  i— 1 1—  co  ci  o  —  r- 

o  ;o  i—  i^-  x  x  ci  o  o 

o  in  >n  o  m  in  >n  ~~c  so 

XXXX  XXXXX 


O  X  O  CO  Cl  -t  X  rH  -f  iC  O 

ci  o  -M  -t  m  i-  x  o  i-i  CM  co 

O  t-  CO  Cl  O  T-I  I-  -t  O  «D  CM  j 

;O  JO  1-—  1^-  OO  Gl  Ci  O  rH  -^  CM  C? 

XXXXX  XXXOOX  X  X  CO 


1^  X  X  Ci 
CO       CM  X  -t  O  O 

-t     m  in  :c  i-  i— 


-  X  Ci  X  1- 


in  in  in  m  o 

XXXXX       XXXXX 


X  <£>  -t  rH  t-       CMOOCM-t       01 

*>**    S^gS    SI 


GO  -t 
O  O  I—  i—  X 

X  X  00  00  X 


_S    ££ 

££   c2J2 


OCMt 1  -t  CDXC1X1- 

r-t   CM   CM  CO   CC  CO  CO  CO  CO  CO 

(M  GO  -t  O  O  CMX-tOSC 

m  o  ;c  i-  i^-  x  x  ci  o  o 

m  in  m  m  m  in  o  in  ^  ^ 

XXXXX  XXXXX 


'-t     "^t     CO     1— '      Ci 

i-l  CM  CO  't  -t 

^N  ^^  KJ     i— ( t^*  co  ci  in 

OOi-(       CMCMCOCOrfi 

m  in  in 


ci  o  i— i  in  ci  i— i 

-H  co  m  so  i—  ci 

X  -*  O  SO  CM  X 

t-  t-  OO  Cl  Ci  " 

OOXX  XXXOOX       XOOXOOX 


O  **       O  X  Cl  X  X 


•^  CO  —"  X  -* 
O  CM  -t  iO  t- 

in  co  t—  t— 

X  X  X  X  00 


t—  O  CM       COCOCMi-HX 
^HCO^        ia  ®  t-  GO  GO 
Tt<       O  SO  CM  X 


O  CO       O  X  X  00  X 


CM-HC1^>CM       GO  <N  O  Ci  *-!       CM  CM— 'OX       -tOmOCO       Oi^-XClX 

-t^t-cii-H     cM-tin^x     cio-^cMCM     co-*-tmin     minoom 


in  —i  i-  co     ci^DCMco-t 


^^     *U1     ^     l^    U'J  ^*     ^     :-.^     'Vv     ^1  ^^     •'^     S»N 

xciooi-i  —(CMcoco'*  inocc 

•*t  -t  in  m  in  o  in  m  o  in  m  in  m 

XXXXX  XXXXX 


iO  *-O  *O  i-"7 

do  cc  x  x  x 


I-  CO  O  O  SM       X 

in  o  i—  i—  o( 

oo  x  x  x  a 


i-l  m  ^  O         rH  r-l  O  Cl 

o  x  ci  o  CM     co  -t  o  in 


•**  o  in  ci  co  in  i^  x  ci  x 

£.  oo  x  x  ci  ci  ci  ci  ci  cr. 

Cl   in  r-l  1-  CO  Ci  O  rH  t-  CO 

OOCiOOi-i       i-HCMCOCO-t*       •*#  *O  SO  SO  I--  t—  X  Cl  Cl  O 

Tt<  -t  in  in  10     in  in  in  in  in     in  in  o  in  in  in  in  in  in  cc 

XXXXX   XXXXX   XXXXX  XXXXX 


-co     ci  in  rH  i~-  co 


cor^incoci  ocico^x  oooxo  coci-tcicM 

i-i  co  o  i—  x  o  r-  co  -t  in  i—  x  ci  ci  o  1-1  1-1  CM  CM  co 

i-  co  ci  in  I-H  x  -<t  o  so  CM  x  ^t  o  so  co  ci  in  rH  t-  co 

in^oi-x  xciooi— i  rHCMcoco-t  ^moot—     i—  x  ci  _ 

-*  -p  **  -t  *t  ^t  ^  in  in  in  in  in  in  in  in  in  in  in  in  in     in  in  in  in 

X  00  X  00  00  X  OO  X  X  X  X  X  X  X  OO  X  00  X  X  X       X  X  X  X 


>  t—  x  Ci  x 
co  co  co  co 


H    «    M   ^    MJ        ?Or-X®O        HNCO^IWS        Or»«COSO        H 
OOOCO        OOOOH         HHHHH         ^r^i-lTHW        W 


LOGARITHMS. 


O  O  O  O  Ci 

CD  CD  CD  CD  o 


Ci  Ci  Ci  Ci  O      O5  Ci  Ci  Ci 


§   §§SSS   88gg£ 


5i 


CD  to  o  o  CD   CD 
X  CO  CO  GO  CO   CO 


'M  »O  t-  00  Ci 
»O  ^t*  CO  <M  TH 
"  CD  <M  GO 

»O  CD  CD 


Ci  CO  O  CO  O 
O  Ci  CO  £"•  CD 

CO  CO  CO  GO  GO 


§O  iC  GO  ^H 
CO  TH  Ci  00 
Ci  lO  O  CD 
O  TH  C<1  <N 

00  CO  00  00  00 


<N  ^*  "* 

co-^cs 

<M  00  "<t 

CC  CO  "^ 


£i^ 


O  <M  GO  -t  Ci 

TH  TH  O  O  Ci 

-t  O  CD  '-N  t— 
TH  ^  -N  CO  CO 


GO  Ci  O 
I—  O  CO 
.0  TH  I- 

lO  CD  CD 


;b 
00 


t^  Jt^t-t^r-t- 

GO   00  GO  GO  Gfc  CO 


t-  t^  GO  Ci  Ci 

GO  GO  GO  GO  GO 


CO  CO  GO  *~^  "^ 
C^  TH  Ci  GO  CD 
GN  00  CO  Ci  O 
O  O  TH  'H 

0000000000   GO  GO  00  00  00 


GO  00  00  GO  GO   GO  GO  GO  GO 


CD  CO  O 


CD  CD  CD  CD 
GO  GO  OO  GO 


,  .  _  Ci  t-  CD 
TH  t-  <N  GO  "* 

t-  t-  GO  GO  OS 


TH  rH  (?^ 


(M  CO  •«*  Tj^  CO 


OOQOGOOOGO   CO  CO  00  00  GO 


CD  t  O  CD  TH 


Ci-*GOTH  COlOCDCD 
t-CO^CO  i-(  OS  t-  O 
§^^§?  Ci^QCC 


0000  00  00 


t—  GO  CO  t- 

CO  TH  Ci  t- 
^  O  »O  -H 

OOOOOOCO        GOOOOOOOGO 


Ci  t- 

G<l   t—  CO  Ci  iO  TH 

<N   C^  CO  CO  ^  iC 

GO  GO  00  CO   GO  00  GO  GO  X 


'tTHCOCO  t—  TH  ^  CO  t-  GOt'-CO^iM  OO^CiCOCD  OS  -H  (N  <N 
COCO<M<M  THTHOCiGO  t-CD»O'TCO  TH  O  00  t—  O  COC<IOOC 
OTHr-cO  CiOTHCD^  GO^OCDr,  CO  g  g  JO  TH  ^COC,^ 


GO  GO  00  QO   00  GO  GO  GO  GO 


i6 


LOGARITHMS* 


X  GO  r-  iO  3<J 

os  CD  cc  o  r- 

00  TjH  O  CD  -r-l 

TH  cq  co  so  ^ 

00  CO  00  GO  OO 
GO  GO  OO  00  GO 


OS  lO  O  -*  GO 

co  o  r-  co  os 
t-  co  oo  •<*  os 

HH  »O  iO  CD  CD 
GO  00  GO  GO  CO 
OO  00  OO  GO  GO 


C<1  O  CO  <N  t- 

O  GO  O  CO  O 
i-i  CO  CO  X  ^* 
CD  tO  t"*  t—  GO 

GO  GO  00  00  00 


TH  lO  .t—  OS  TH 

co  o  co  os  r- 

OS  lO  TH  CO  <M 


OO  Tt<  OS  lO  TH 
TH  CO  CO  CO  •* 
OO  00  GO  CO  GO 
00  00  00  00  CO 


CO  00  CO  GO  CO 

OO  "*  i— I  t-  ~* 

CO  CO  X  CO  OS 
-tf  O  >O  CO  ~ 
GO  GO  00  00 
GO  GO  00  OO 


i  co  os  HH  cs 

CO  OS  t-  -* 
>  CO  TH  t-  CO 
CO  t- t-  00 


^  t—  O  CO  CO 

c5i  os  t-  ^  TH 

OS  -t  O  CO  CO 

'X  OS  O  O  TH 

r-  t—  GO  oo  GO 

00  GO  GO  00  CO 


Tf<  TH  CO  TH  OS 

00  »C  CO  OS  lO 

i-  co  OS  HH  o 

TH  (N  <M  CO  HH 
OO  GO  GO  GO  GO 
OO  OO  GO  00  00 


»C  CM  t-  1-1  O 

CM  OS  iO  <M  GO 

<r>  TH  t-  co  ~ 

"*  O  O  CO 

GO  00  GO  00  GO 
OO  'GO  GO  00  00 


OS  rH  CO  "•*  ^ 
HH  ,H  t-  CO  OS 

•«*  O  10  TH  -- 


xxxxx     xxxx 


CO  O  CO  iO  CO 
CO  ^  TH  GO  O 
00  -<t  O  if5  TH 

GO  OS  O  O  TH 
t-  t-  GO  00  GO 
CO 


TH  CN  <M  CO  -^ 

CO  GO  00  X  00 
CO  QO  GO  00  00 


O  O  O  *O  OS 
CO  CO  O  CO  <M 
»O  TH  1—  (M  GO 
•^  »O  »O  CD  CO 
GO  00  00  00  00 
GO  00  00  00  00 


<M  lO  CO  t-  00 
OS  lO  TH  1^*  CO 
CO  OS  iO  O  CO 
lr*  IT*  00  OS  OS 

GO  00  OO  GO  OO 
00  00  OO  CO  GO 


§r~  co  os  ^ 
O  GO  ^O  CO 
o  o  co  rN 

»O  CD  t-  1—  GO 

t- 1-  r~  t-  r- 

00  GO  CO  00  GO 


OS  CO  CO  GO  OS 
O  GO  O  <M  OS 
00  CO  OS  O  O 
GO  OS  OS  O  TH 
t-  1—  1—  GO  00 
00  00  00  CO  00 


o  o  cs  r-  o 

t—  H^  O  lr-  Tj< 

CO  CM  CO  CO  OS 
TH  <M  <M  CO 


<M  00  rf  CS  CO 

TH    t—    Tt    O    £- 

»re  o  co  <N  t- 

-  O  O  CD  CO 
'00  GO  00  00 
GO  00  00  CO 


CO  00  O  TH  C<J 
CO  OS  CO  CVI  GO 

CO  CO  "st  O  >O 

CO  OS  OS 


rH  OS  CO  <M  t- 


00  ^*  O  CO  TH 
O  CO  t-  t-  GO 

CO  00  OO  OO  00 


CO  iO  X  TH  CO 
O  CO  OS  1—  H/l 
t-  CO  X  •«*  O 

X  OS  OS  O   rH 

t-  t-  t-  X  X 

xxxxx 


CO  CO  <N  TH  00 

TH  00  »O  fN  00 

co  1-1  t—  co  co 

TH  <M  <M  CO  CO 

OO  OO  GO  GO  OO 
00  00  GO  GO  GO 


iO  C^  t-  <M  CO 

O  ^  00  »O  TH 
HH  O  O  rH  t- 

Tj<  O  iO  CD  CO 
00  00  GO  GO 
00  00  GO  00 


00 
OO 


—  ,  _  CO  CO 
•CO  X  Tt<  OS  >O 
t-  t-  X  X  OS 

xxxxx 
xxxxx 


CO  TH  X  -^  OS 
TH  OS  CO  "<*  TH 
X  CO  CS  uT5  TH 
»O  CO  CO  t—  X 

X  X  X  X  X 


"*  X  TH  CO  iO 
OS  CO  ^*  TH  X 

CO  (N  X  HH  OS 
X'  OS  OS  O  O 

x 

X  X  X  X  X 


CO  CO 
iO  <M  OS  CO  CO 
»O  TH  CO  CM  X 
TH  <M  -^  CO  CO 
X  X  X  X  X 
X  X  X  X  X 


OS  lO  TH  CO  O 

os  co  co  os  co 

CO  OS  iO  O  CO 

TjH    Tj*    O    CD    CO 

X  X  X  X  X 
X  X  X  X  X 


t—  co  os  TJH 

r- 1-  x  x  os 

X  X  X  X  X 
X  X  X  X  X 


X  X  X  X  X 


<M  OS  -*  OS  H^ 

CO  OS  HH  O  CO 
Hi  H^  lO  CO  CO 
X  X  X  X  X 
X  X  X  X  X 


OCO  "* 
lO    TH 

TH  t-  <N  X  ^* 

t-  t-  X  X  OS 
X  X  X  X  X 
X  X  X  X  X 


X  X  X  X  X 


O  T<I  X  CO  1— 

X  »O  TH  X  ^ 
•M  X  ^  OS  O 
•^  H^  to  »O  CO 

X  X  X  X  X 
X  X  X  X  X 


TH    T*    CO    1—    X 

TH  i-  co  os  o 

TH  CO  (M  1"-  CO 
£-  t—  X  X  OS 

X  X  X  X  X 
X  X  X  X  X 


cq  co  os  (N  ^ 


X  X  X  X  X 


X  O  CO  OS  CD 
CO  OS  iO  O  CO 

X  X  X  GO  X 
X  X  X  X  X 


OS  >O  TH  CO  -H 
CM  OS  CO  Ol  O 

<N  r-  cc  as  -ti 


»O  TH  t—  H/t  O 

O  CD  TH  t-  CO 


xxxxx     xxxxx 


wo      « ® *  §  * 


LOGARITHMS. 


117 


SSSSS  S££88  £88£8  £88£8 


O  »0  iO  »0  »C 


CD  i-i  ^  IT-       O  1-1 
O  «C  i-H  «O       -N  t- 

<>i 


O  »O  O  -*  t-  OS  -T-I  (N  <N  <M  i-lCSCOCOOS  ID  O  •<*  i—  O 
"^OSiCOO  O  CD  1-1  O  1-1  COOiOO^  OS^GOCNt^ 
O>OrHlr-G<J  GOCOCS-fO  lOi-HCDC^t-  CNOOCOOS-^t 


O  «O  i-H  >O  GO  i-t 
T^  GO  CO  t- I-H  O 
O  rH  t^  CM  00  CO 


CO  OS  "^  GO 
t^  'N  00  CO 

CO'tOSiO 
(MCOCO-^ 

os  os  o  os 


OO       00  GO  GO  GO 


i-H  CO  i-H  -^  t- 
CO  t-  (N  CD  O 
I-H  ?D  (N  I—  CO 
I-H  T-I  (M  <^  CO 

O5  OS  OS  OS  OS 


•  '^^  ut^ 

OS  •* 

00  GO  GO       GO 


'-IOSO  (N  GO  CO  t—  O  CO>£?COI:-1— 

OS  "^  O  CO  i^  1^  <^l  GO  CO  GO  CO  GO  CO 

Ot  ^~-  Ct  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS 

CO  GO  CO  COOOGOOCGO  GOOOOOGOOO 


(N  t-  (N  »C  OS 

CO  T-I  »O  OS 

CO   ^H 

<N  CO 
11 


GO  t"*  ^O  CO  O  f'"  C^  1^*  i^  *O  GO 

OiCi-H  i-'Soocoos  -^o»o^o      o^^c; 

os  os  os  os  os  os  os  os  os  os  os  os  os 

00  OO  00  00  GO  GO  GO  GO  00  GO  00  00  GO 


1—  >N  CO  O  CO»OCOt-t-  COKt>OSi~it-  COCOCOt-O 
OCOi-Ht-  (Mi—  S<l  t-  CM  l—'Mi—  iMCO  THiQO^t-CS 
-Nt-COOO  "^OSOOCO  i-il—CNCOCO  OS-tOOC 


gliil  lilll  Illii 


LOGARITHMS. 


"•*  -*  -*  ^  ^ 


»o  »o  »o  o  us     o  »o  o  o  o     S  S$  »o  »o  §     o  ic5  »b  o  ib 


O  i-l  CM  CM  i-H 
CMCOO-*  GO 
^  t- 


t— i-i^tit^  osoi-ii-io  ost^io^r- 

t^OCO  CO  O  CO  CO  Oi  -rHTtHt-OCM 

CO  — 1  t-  CM  1—  CM  COCOCO^OS 

OrH— (  CMISCOCO^*  ^  iO  »O  CO  CO 

OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS 


OS^COi-l^t<  COt-COGCL—  CO-*CMOS»C 

"~   -H  »O  CO  i-l  -t1  1—  O  CO  CO  OS  CM  -t  1~- 

>OOO  i-HCO^Hi-CM  t-  CM  GO  CO  GC 

OI-HI-H  CMCMOOCO't  -tiOiOCOCC 

OSOSOS  OSOSOSOSOS  OSOSOSOSOS 


O  i-l  CN  (?q  CM 


co  o  co     I-H  co  I-H 

t-  i-l  -^         CO— liOGOrH 

»o  1-1  co     1-1 


os  os  os      os  os  os  os  os      osososcsos 


COi-HOSCOCN        GOCOt-i-H^       t-GOOSOO 
Tjn  CO  1-1  »O  OS        CM'-OOSCOCO       OSC^OOSCM 

^os»ooo         « 


CO  OS 

CO  <C 

3  £  gig 


os  os  os  os  os  os  os  os  os  os 


OS  t—  O  CM  OS  -tO^COi-t 

COCMOOCO  r-~-<-'tJt^i-i 

cBcOt-GCCO  CSOSOOi-i 

OS  Gs  OS  OS  OS  OS  OS  OS  OS  OS 


-^  t—  o  co 

r-  CM  GO  CO 

^(MCNICO 


co  10  co  o  t- 

OS  (N  iO  CO  O 

co  os  "*  os  >o 

rt  -rt  »O  >O  CO 


OS  OS  OS  OS  OS   OS  OS  OS  OS  OS 


S^IiOCOi— I  rtit-OCMO 

i-H        CO(Mt^(MCO  COCO-^OS"<* 

— i        i-l<MCMCOCO  "^-^lOiOCO 

OS  OS  S    OS  OS  OS  OS  OS  OS  OS  OS  OS  OS 


O       H   N   W  ^  «5 

00       00  «  00  00  00 


CO  r-  X  05 

So  So  So  So 


H   N   CO 


LOGARITHMS. 


119 


fi  § 


CO  X  71  CO  Ci  —  CO  -i*  -t  rJH  -*f  71  O  X  "*  -H  CO  TH  >O  OS  CM  iO  t-  X  Ci 

iO  1—  O  71  -t  I-  Ci  —  CO  »O  t-  Ci  T-I  CM  -t  CCt-CiO-H  CO-^iOCCI^ 

-t  Ci  iO  O  iO  O  »C  —  CD  — t  CD  TH  1—  71  t-  71  t-  CM  X  CO  X  CO  X  CO  X 

1-  L—  X  Ci  Ci  O  O  i—  TH  CI  71  CO  CC  -t  ~f  iO  »O  CO  CO  t-  t-  X  X  Oi  Oi 

—  T-H  T-I  —  — i  7-1  71  71   71  71  71  71  71  71  71  71  71  71  CM  CI  71  71  71  71  71 

CiCiCiCiCi  CiCiCiCiCi  CiCiCiCiCi  CiCiCiOiCi  OiCiCiOiCi 


i  O  iO  O  CC  CD       Ci  TH  71  CM  7}       <M  O  X  CD  CO  Ci  >O  O  -t<  X  TH  CO  iO  I—  t- 

O  CI  iO  t-  Ci       T-  -t  CD  X  O       CI  -T  iO  I—  Ci  O  71  —  O  CD  X  Oi  O  T-H  CM 

ar       -f  Ci  -t  Ci  -t        O  >C   O  »C  — i        CO  T-H  CO  T-H  CD  CM  1—  CM  1—  CM  t-  CM  X  CO  X 

i  I—  r-  X  X  Ci       O  O  —  —  71       71  CO  CO  -F  -*  lOOCDCOt-  t-  X  X  Ci  Ci 

CI  CI  CI  CI  CI  71  71  CM  CM  CM  71  71  71  <N  CM 


x  cc  r-  i-i  -t 

-t  t-  Ci  71  -t 
CC  CO  CO  Ci  -t 
I—  1—  X  X  Ci 


Ci  Ci  0  Ci  Ci 


i—  x  _  is  -T       us  i—  BU 

|OCDT-,CO        THCOTH 


~  71  71  71  7-1        C^CMCMCMCM        CM  7}  CM  71  71        CMCMT-1CMCM 

Ci  Ci  Ci  Ci  Ci      Ci  Ci  Ci  Ci  Ci      Ci  Oi  Oi  Oi  Oi      Oi  Oi  Ci  Ci  Ci 


iCO»CCi71  -fCOXXX  Xr-»OCMCi  OT-I|^T-H»O  XT-HCCIO»O 

Ci  71  -t   ~   Ci  —  CC  iC  I-  Ci  T-  70  i-C  I-  X  O  71  CC  1C  CD  t—  Ci  O  T-H  CM 

T1XCOXCC  Ci-t<Ci-fCi  iCCiCOiC  i—CD-HCDrH  CD  I-H  i—  CM  f— 

t—  t-  X  X  Ci  Ci  O  O  T-H  -H  C-l  CO  CO  -t  -f  uC  iC  CD  CD  i-  1-  X  X  Ci  Ci 

,_,  ,_,  T-I  T-I  _  T-  71  71  71  71  71   CI  CI  CI  d  CM  CI  71  71  71  71  71  71  71  71 

CiCiOiOiCi  CiCiCiCiCi  CiCiCiCiCi  CiCiCiCiCi  CiCiCiCiCi 


^  CD  Ci  T-I  -<t 
71  t-  71  X  CO 
i-  i—  X  X  Ci 

Ci  Ci  Ci  Ci  Ci 


-t  O 
X  O  ' 


iC  CC  T-H  X       rf  O  »O  O  "*       1—  O  71  CO 
X  O  7J  CO        iC  I-  X  O  — i        CM  -T  O  CO 

OT-HT-H       CMCMCO^1^1       lOiCCOCOr-       F^XXCi 

-1  71  71  71 


33333   3333 


OTlCC-f4'*  ^COi-HCiCO 

0>  T-H  "*  CD  X       T-  CC  iC  1-  Ci  T-  CC  iC  CO  X 

T-I  1~  71  1—  71        X  CC  X  CO  X  "^  Ci  -t  Ci  "* 

t-  t—  X  X  Ci       Ci  O  O  T-H  -H  7-1  71  CO  CO  Tf       iO  iC  —  CD  1- 

TH  T-H  T-I  T—  T-H        —  CI  71  71  71  71  71  71  71  CM        71  71  71  71  71 

Ci  Ci  Ci  Ci  Ci      Ci  Ci  Ci  Ci  Ci  Oi  Ci  Oi  Ci  Ci      Oi  Ci  Ci  Ci  Ci 


COX^XCM       CDX-HT1CO 
O  T—  CO  ^  CD       i—  X  O  T-H  71 

0,00,00       .OOCOT.CO 


Ci  Ci  Ci  Ci  Ci 


T-I  O  t-  "*  T-I  1—  CM  I—  -H  -ft—  Ci  T-I  71 

x  ci  T—  co  icccxcii-i  coco-rcoi— 

X  CO  Ci  "^  Ci  ""^"  Oi  "^  O  iO  O  iC  O  iC 

71  71  71  71  71  71  71  71  71  7~1  71  71  71 

Ci  Ci  Ci  Ci  Ci  Oi  Ci  Oi  Ci  Ci'  Ci  Oi  Ci 


§1— i  CO  O  X  O  CM  -*t  t—  Ci  T-H  71  —f  iO   J 

'*£.  ~  t£  1-4  t-  71  i"-  71  t—  CO  X  CO  X  0 

1—  t—  X  X  Ci  Ci  O  O  — i  TH  T-ITICOCO^ 

TH  TH  T-I  —  TH  TH71717171  7I7171CM7 

Ci  Ci  Ci  Ci  Ci  Ci  Ci  Ci  Ci  Ci  Ci  Ci  Ci  Ci  Oi      Oi 


Ci     Tt"     Oi  ^t1      Ci      "^     O     1C 

iO  CD  CD  t-  I-  X  Ci  Ci 

50  7^  71  CO  71  CM  <71  CM 

OiCiOi  CiCiOiCiCi 


CO  X  CC  t—  O       CC  CO  t-  X  CO       Xt-CO-*rH       X-^Oi^X  <M>Ot-CiO 

O  CO       iC  I"-  Ci  TH  CO       »O  t-  Ci  T-H  CO       Tjn  CD  i—  Ci  O  CM  CO  'f  O  I— 

COT-H       CDT-^CDTli—       CM  t—  71  X  CO       XCOXCOCi  •^Oi^tCi-t 

lOO^-^H        CMCMCOCO-^       TfijCiCCDCD  r-  t—  CO  00  OS 

CM  71  CM  CM  71  CM  71  71  7^1  71 

Ci  Oi  Ci  OS  Oi  Oi  Oi  Oi  Ci  Ci 


l 
0  0 


Ci  Ci  Ci  CT.  Ci 


O  CO  O  >C  X  T—  CC  iC 

X  O  CC  iC  I-  O  CM  -t 

Ci  iC  O  «C  O  CD  -H  to 

CD  I  —  X  X  Ci  Ci  O  Q 

Ci  Ci  Ci  Ci  Ci 


»O-*CMCi       OTOXCMt—       OCOCDXCi 
Cl-tCOt"—       Cii— iTl-^iO       Ir— XCiO-H 

C^  CM  CM  71 


_.        _.        _.       ..  _.       _.       _.        _.        _. 


O  r-  X  O 

«   «   «   « 

00   X   X   00 


n  •<*"  w 

n  co  w 

X  X  X 


O  *-  X  O 

ct  x  x  x 


I2O 


LOGARITHMS. 


cs  x  t-  o  co     o  r~  co  x  co 

X  OS  O  ri  <N       CO  00  -t  -t  O 
OS  ^       OS  -t  OS  -t  05 


CO  X  "* 


r^ococox  cs  o  o  os  co  t-^oix>o 

»(7  co  co  ;c  co  co  i  -  i—  co  co  co  co  cc  o  o 

-f  CS  -t  CS  -t<  OS  -t  CS  -f  OS  -tOS-tCS-f 

uO  O  CC  CC  I—  i—  X  X  OS  OS  O  O  — i  I-H 


^  ZZ 


OS  OS   OS 


C5  C5 


X  1—  CO  iO  <M 
CO  ^  O  CO  1- 

CO  X  CO  X  CO   «u  >.w  --. 

o  o  r-i  T-I  rM      ,ncoc< 
co  co  co  co  co 


OCOCOX       OSOOCSX       ^OOl 


O  O  CO  CO  1--       1--  X  X  OS  OS 

co  co  co  co  co     co  co  co  co  cc 

O5OSOSOSOS         O5C5OSCSO 


OS  »(0   rH  1^  <M 
r-    <N  (N    CO   Ttl    Tjl    1C 

x  co     x  co  x  co  y. 

"H  rH   Ol         (N   CO  « 

co  co  co     co  co  c<v  ...  .w 

OS  OS  OS       OS  OS  OS  OS  OS 


CO  lO  1—       OS  O  O  O  OS       Jt—  O  CO  CS  CC 

"isti 


Rococo     co 

os  cs  os     os 


O  TH  oq      01  co  -t  — f  o 

iil 


P  P  3     rr  o  co  o  cp 


cocci---     r^xxxos 
ososos     os  os  os  os  oi 


ct 


CO  CO  rH  OS 
CO  rH  CO  rH 


O  C5  Ol  »C  t- 


X  O  CO  O  t- 


CC  rH  5O  rH  t— 

(M  CO  CO  ^f  ~tl 
CO  CO  CO  CO  CO 
OS  OS  OS  OS  OS 


<M  t—  Ol  t—  oq      IT—  55  t—  oq  1--       (M  lr—  <>l  t—  CM 


oococcr—     i^-xx 
co  co  co  co  co      co  co  co 

OS  OS  OS  OS  OS    OSOSOS 


OS    O  O  r-i  rH 
CO    T^  ^t1  ^  TjH 

OS   OS  OS  OS  CS  OS 


C<1  CO  CO  rt< 

O  rH  CO  1-1  CO 
<M  CO  CO  "<t  "t 

CO  CO  CO  CO  C>3 
OS  OS  C5  C5  OS 


**  OO  -M  lO  b- 

10  10  co  cc  co 

rH  CO  rH  CC  rH 

iO  O  O  O  t- 

co  co  co  co  co 

CS  OS  OS  OS  OS 


coo0-0- 
co  i-t  co  i-i 

£-  00  X'  OS 

co  co  co  co  co 
os  os  os  os  os 


<M  (M  rH  O  CO 

co  co  co  co  co 


t-  X  X  OS  OS 
JO  O  »O  O  O 
GSI  CO  CO  ^  -* 
CO  CO  CO  CO  CO 
OS  OS  OS  OS  OS 


X  OS  O  O  OS 

i-l  r-,  rM  CM  ^H 

co  -H  co  ^  co 
i-  X  <X  OS  OS 

co  co  co  co  co 

OS  OS  OS  OS  OS 


00  CO  -^  T-H  t- 

T-I  r-,  ^  TH  O 
i-H  CO  1-1  CO  i-l 

O  O  T-<  -^  (M 

-t  -t  -r  ^  Tt 

OS  OS  OS  OS  OS 


-^  -H  GO  TjH  OS 

cq  co  co  ^*  TJ< 

ift  O  lO  O  >O 
G<l  CO  CO  -t1  ~f 

CO  CO  CO  CO  CO 

OS  OS  O  OS  OS 


C  O  CO  CO  t- 

ro  co  co  co  co 

OS  OS  OS  OS  OS 


CS  OS  OS  OS  OS 


O  OS  GO  CO 

Tfl  ^  1O  CO 

^  OS  Tf  OS 

CO  CO  CO  CO 


^  rH  t>  CO  X 

i-  X  X  OS  OS 
^f  OS  -<t  OS  Tj< 
<M  3^1  CO  CO  ->t 

co  co  co  co  co 

OS  OS  OS  OS  OS 


CO  t-  rH  -^  CO 

O  O  rH  T— I  T— I 

CO  CO  CO  CO  CO 
OS  OS  OS  OS  OS 


XOSOOOS  COCC^T-lX 

rHrHfN<MrH  rHrHrH-HO 

>O  O  O  O  O  O  O  O  »-O 

r-XXC535  OOrHrH 


O      H    N    W  ^J  10 

90      00  00   00  00  00 


LOGARITHMS. 


121 


Q"       O  <O 
OO 


.  .  CO  I-H  CS   CO  CO  CS  »O  O   •<*  GO  TH  ^H  i- 
GO  CO  X  CO  GO   CO  CO  C*  t-  (N   £-  71  I-  ^H  CO 

os  os  os  os  cs  cs  cs  os 


I   1-1  O  O  -*  t-  O'N-tiO^) 

O  OS  OS  C^  1?^"  I'*  CO  iO  "^1"  CO 

OS  CO  OD  CO  QO  CO  GO  CO  00  CO 

•M  CO  CO  ^  T)H  »O  »O  CO  O  t— 

33355S  S3BSSS 


CD»O^(MO  t- 

C^  T-H  O  OS  OO  CO 

CO  CO  00  C~l  1—  <M 

r—  CO  CC  OS  OS  O 


-t<  O  CO  T-H 
.0  -t  ^  ^H 

t-  (N  t— 


T-H  co  IN  o  GO  ~-i  co  10  co  r— 
•5  t  -?  w  7?  09  ?H  Q  os  co 


r-  co  o  -*  co  in  o  os  t-  co  -*t 

i—  (M  t—  (M  t-  <N  1^  i-H  CO  rH  CO 

'      XGOOSOS  O  O  ~H  — 'iN  <N 

^  "^  ^  "^  iO  *O  tO  *O  iO  iO 


cs  os  os  os  os 


iH  t>.  T-(  O  O5 


(N  -t 
t-  CO 
<Nr-<Ml^iCVJ 

OiOCOCOi— 


COt-COiOCO       Ot^COOS'*        OS 
t-JO^'NrH       Ct 


^Sq^Sco     t-.co^S^ 

t—  X  X  OS  OS        ~    ~ 


osososos      osoiososos 


I-  71  I-  7}  1—       (N  i—  M  CO  T-H 

71  CO  CO  -t  -t        iO  O  CO  CO  I— 

OS  OS  CS  OS  OS       C5  ct  OS  ct  3! 


I—       1—  X  X  OS  OS       O  O  —  "-»  <M       (M  CO  CO 
>-t*        rt  T^  ^  Tt<  -f       »O  >O  »O  O  »O        O  »O  O 

OS        OSOSCSOSOS        CSCSCSCSO5        CSCSOS 


fMGO'MCOO  CO  »O  t—  OS  O  OCS 

22SiSSg  SiS33  S^cooo  000.60 

71  CO  CO  -t  ^  >O  O  CO  CO  1—  t-  X  X  CS  CS  O  O  1-1  i-t  iN 

•^  Tf^  r^  -^  -^  ^  -^  -^  -^  -^  — ^H  --^  -^  --^  T^  iO  i^  iO  *O  »Q 

oscsoscso  oscsososos  cscscsoscs  cscscsoscs 


.      .    ____  .  .  .       _ 

—  i  CO  1-1  CO  r-iCO-^ 

CO  CO  Tt<  "*  O  iO  CO 

cs  cs  os  cs  CSCSGSOSCS 


O       TH  O  O  X  CO 

CS       X  I"-  CO  -t  CO 

o     »o  o  o  o  «o 

t—  X  X  OS  OS 


os 


C^7,§^ 

2:2  3*3x3 


-t1  '-i  t—  CO 
7-1  I-H  OS  X 

§S8^-H 

o  o  o  >c  o 

cs  c:  cs  os  cs      cs 


(N  (M  — '  CS  X   O  71  Ci  »O  O  O  OS  CO  CO  OS 

71  -H  OS  X   t-  CO  "!f  CO  -N  O  X  I-  O  CO 

O  10  OS  -f   CS  -f  CS  -*  Ci  ^  GO  CO  X  CO 

-  i-t  i-H  CM  74  CO  CO  -t 


-*OS-HGO(N  lOCOO^H 

»o^t-tcoco  7^1—1^-0 

O  O  iO  O  O  O  >O  O  »O 
(N  o^  CO  "^f  "^ 

cscsososcs  cscsoos 


OCSCO       CO  OS  — i  <N  CO 
OSCOX       t-COCO»O"<t 

7^  CO  CO  CO  "^        "^  »O  O  CO  CO 


*~  OC  C5 

QC  X  X 

00  00  00 


122 


LOGARITHMS. 


CO  GO  CO  CO  CO        CO  CO  CO  CO  CO 


333 


GO  O  O  i-H  O  CI  X  CO  ^  T-H  Jr-  CO  OI 
to  ^  <M  ©  CO  tO  CO  i-  Ci  1-  -f  -N  01 
rH  CO  TH  to  O  O  p  O  Ci  -t  Ci  -t  X 


nil 


Jr-  CO  Oi  -t<  X        (M>OCOrH'M 

Ci  i — h  ^  ci  o  -^  — i 

Ci  -t  X  CO  'X   CO  l~  -M  l~  OJ 


O  <M  fN  CO  <N  <MO 

T-H  Cl  i—  O  CO  i-1  Ci 

i— i  to  O  »O  O  O  Ci  "t  Oi  -t< 

»0  O  CO  CO  i-  1  -  1-  X  X  Ci 


OS  OS  OS  OS  OS 


CO    rH    CO    i-H  JO   X   rH   CO   >O 

t"*  JO  CN  O  1^~  ^f  ^  Ci  CO 

CO  X  CO  X  (M  l~  C^l  CO  T-H 

11 I I  ilili 


t-COCi-t 

O  CO  O  X  O   CO  O  1~  »i?  <N 

-  ^  1-  1-1  O   T-H  ^  O  »<7  O 


-H  — i   fM  -N  CO  CO 

OC;   O  ^  ^  O 
Ci  Ci   Ci  Ci  Ci  Ci  Ci 


(M  CO  "*  iO  "* 

CO  -t<  TM  ©  X 
O  »O  O  iO  Ci 


-H  X  »O       (N  X  't  Ci 


i—  i—  co  co  Ci 

»O  >O  1C  O  >O 

Ci  Ci  Ci  Ci  Ci 


i  CO   1^  1-1  -*  ?C  CO   O  i-<  TH  T-H  O 
-  »O   CM  O  i—  -*  i-t   Ci  ^  CO  O  I- 


X  CO  X  CN  l-^  M  1-  T-H  CO  -H 

Oi  O  O  i-t  T- t  (M  OJ  CO  CO  't 

,0  CO  CO  CO  CO  CO  -O  CO  CO  CO 

Ci     Ci     Ci     Ci     Ci  Ci     Ci     Ci     CS     Ci 


T-H  GS  i—  10  oo     TH  ci  r- 

O^Oi-*Ci       Tt^X-CO 


Ci  Ci  Ci  Ci  Ci   Ci  Ci  Ci  Ci  Ci 


CO  — •  CO  OCOCOCi-H 

iO  CO  O  X  iO  fM  Ci  t— 

t-  ^  t_  ^H  ;o  T-H  iO  O 

»"ococococo  cocoiococo 

OiOiCiCSCi  OCi'CiCiCi 


O 
<N 


'Cit-O<M 
OS  I-  ^  CO 
CiCOXCOX       COl^-CNt—  CN 

^OiOOCD       1—  I—  XXCS 

•o  o  >o  10  10      >o  >ra  io  10  o 

Ci     Ci     Ci     Ci     Ci  Oi     Ci      Ci     Ci     Ci 


— i   >O  Ci  ^N 


Oi  TH  5<1  CO  CO 

CO  lO  CO  rH  Oi 

X  CO  X  CO  t— 

JO  >O  *O  iO  iO 

Oi  Ci  Ci  Ci  Ci 


(70  -H  Ci  t—  >O  T— (XCOCiCC 

t-»O(NOX  COCOT-^XCO 

>O  >O  iO  «O  »O  »O  CO  CO  CO  CO 

Oi     Ci     Oi    Oi     Ci  Oi     Ci     Ci     Ci    Oi 


33o<£   3Sii?i 
£000    q;^;5-Se 

K^  to  en  ro        o  ^  tc  CO   ^ 


rH  »0  I-  Ci  i-H  CN  CO  CO 

rH  X   1C   'M         O   t 1   — < 

»O  Ci  -fi  CI  -t  X  CO  X 

co  co  -ti  T*  »ra  »o 


01  fNi  ^a  ^ 

iiiHi 


Tt<  iO  »O  CO  CO       1~  t-  X  X  OS 

iO  iO  >O  iO  O       »O  »O  tO  O  »O 
OS  Oi  Oi  Ci  Oi       Oi  Oi  01  OS  05 


-t  1^  O  <N  Tt»OCOCOCC 
OOO^.(X)  JO^MCiCOCO 
Tf  C^  ^f  X  CO  X  "M  I-  JN 


>0  O  10  »0  O       »0  CD  --0  CO  O 

OiCiCiOiCS       OiCiCiCiCi 


t-  X  Ci  01  Ci 
O  t h  — c  X 

_-      .    „    ..  CO  1"-  "M  t-  T-I 

T-H  (CV|  (CV|  CO  CO         "^  "H^  lO  lO  CO 

occcococo     cococo-occ 

OiCiCiCiCi       CiCiCiCiCi 


__  „  OS  X  t- 

TJH  io  iO  co  co  r~*  t"~ 

o  »o  »o  »o  »o  o  o  »o  »o  »o 

OS  05  OS  05  OS  OS  OS  OS  CS  OS 


iO  01  CO  CO  X 

Oi  CO  Tfi  T-H  X  CO  CO  O 

CO  GO  CO  I—  <N  t-  <M  CO  rH 

CI  Ci  Ci  OI  OI  Ci  Ci  CI  OS  CI 


O       H   N    M 
0)      05  05  05 


to  ?*  x  cs 

05   05  05  05 


LOGARITHMS. 


123 


CO  CO  CO  CO  CO       O  CO  CO  CO  CO       CO  CO  CO  O  CO 


co  — i  cs  co  TO 

TO  O  CO  TO  O 
O  O  CS  -t  CS 
I—  1—  1-  CO  X 


CS  O  O  O  CS  TO  t—  CS  'N  CO  »O  CO  CO  CO  »O  Tfi  CO  O  CO  O 
CO  TO  O  CO  'N  OS  O  — i  CO  -^  O  CO  CM  CO  -*  O  CO  <M  1—  TO 
TO  X  TO  i—  7<l  CO  T-H  CO  O  O  O  -t1  OS  CO  CO  CO  1—  O-l  CO  1-1 


OS  O  O  O  OS       GO  t— 


TO  -t4  -t  >O  iO       CO  CO  1—  I"  CC 


I-H  CO        ^   r-t  1—  TO 
O  ^H         -H   -N  CM  TO 


t- t- 
05  Ci  Ci  O  O 


_     _     -     -     -         1--  t—  t—  t--  r—      1—  t—  I"-  I"-  J— 

OS  OS  OS  CS  CS       OS  CS  CS  OS  CS       OS  OS  CS  OS  OS 


O  ~f  ?<J  CS  CO  CO  CS  -t<  CS 

-t  —  X  T  i-l  CO  -*  i-«  I- 

CO  TO  I-  O1  I—  i-(  CO  1-!  »O 

os  cs  os  cs  cs  os 


i-HTt*l—  OS 

t-  TO  os  o 

OS^COTO 


CM  CO  TJI  O  CO       CM  t-  TO  CS  iO 
COCMt-fNCO        i-iiOO^CS 

os  os  os  os  os  os  cs  cs  cs  os  os  os  os  os  os 


CS  t-  O  TO  O  CD  'N  GO  TO  t-  TH  >O  CO  -H  CO  -t  O  CO  CO  CD 

1  OS  CO  TO  O  1-  CO  O  CO  TO  CS  CO  ON)  'X  >!7  -H  t~  TO  CS  »O  -H 

..   1—  OI  I-  OJ  CO  *-  CO  C '  •-  CS  -t  CS  TO  CO  CO  JC^  7<l  CO  i-H  CO 

CO  t-  1—  X  CO  CSCSOOO  T-H  rH  iM  <M  CO  CO-f-ti>OiO 


»0 


CS  CO 

3s 

X^         SS^°°        rHrHO^OiCO        TO-t-*>OlO        COCbCOt-t^ 

OSCSCSCS       OSC5CSOSOS       OSOSCSOSOS       OSOSOSOSOS       OSOSCSCSOS 


O-l  O  CO  CO  CO 
1^  C^l  CO  ^i  CO 


p  i-H    JO  CS  !N  Tf  CO 

Tf<  OS   Tf^  CO  CO  CO  O^l 
O  O    — H  i-l  IN  !M  CO 


OS  OS  OS  CS  OS   OS  OS  OS  OS  OS   OS  OS  OS  OS  OS 


*M  »O  GO  O 
OS  >O  <M 


»o  o 

-S3 

CS  CS  O  '«^   i"H  ^^  >^  ^.^  t.v   '•'•>  ^r  "^>  ii*1  *i»>   K^  ^^  ^i?  IT"  IT" 

CSOSCSOS       OSOSOSOSOS       OSOSOSOSOS       OSOSCSOSOS 


CO  X  CM  1—  - , 

i-l  i-t  >>l  CM  CO 


CM       O  T^ 


(MCOCSO1'*  Ot^COGOGO 

<NCO'*-Hl—  COCSiOi— (t—  COOS 

CO  t"~  CM  1^*  i^  CO  O  iO  O  "^  OS  ^O 

OO       r-(T-t(Mi7<ITO  CO-^rt<»OO  O^COt^t- 

i^*  !"•     i^*  t^  i^*  t'*  i"*~  i^*  r^*  i^*  i^"  IT*»  j^-  t~^  !"•  i^»  t^— 

CS  CS       05  CS  CS  OS  OS  OSOSOSOSOS  OSOSOSOSOS 


•r-lOGOCOT 


i-HCO       O^HCO-fi-H       i—  "^  O  CO  (N 
CS'tXTOCO        CM  I—  -M  CO  i-( 
CSCSOO        i-ii-IGM'NCO 


co  co  i-  t—     i—  t~-  i—  t—  i—     t-  i—  t—  i~  i—     i^  t^  i^  t—  t- 

cscscsos      ososososcs      cscscsoos      ocsosoos 


O*-OOGSO       i-iNW^UJ        Cl-OOO 
OOCiOO       OOOiOO)       QQQQ 


I24 


LOGARITHMS. 


1—  O  CO  CO  CO 


(N  CO  ^*  O  »O 
<7<1  CO  i— I  CO  O 

CO  O  'O  Oi  "^ 
t-  CO  GO  CO  O5 
GO  GO  CO  GO  GO 

Oi  Oi  Oi  Oi  Oi 


CO  <M  1—  CM  t~ 

-t<  O  »O  .-H  CO 
iO  O  ^  O5  CO 

I—  t-  I—  I—  CO 


T-(  iO  GO  — I  CO 
(N  t—  ;M  GO  CO 

GO  (N  t-  i-H  CO 


co  co  co  -t  ^ 

TJ  CO  CO  CO  CO 

C5  O5  C5  O5  C5 


t-  co  -*  n  o 

co  co  co  co  co 

CO  t-  -M  CO  ^H 
O  »O  CO  CO  t— 


t—  fN  t—  TH  CO 

O  O  -F  C5  CO 
t-  CO  CO  X  O5 

GO  X  'CO  'CO  O) 


(7<l  ^H  Oi  I"-  »O 


OiOiOi       OiOiC5O5O5 


(M  O5  >O  I-H  CO 

co  r-  5<i  t 1 

O  O5  -t  CO  CO 
i—  t-  GO  GO  OS 
CO  CO  CO  CO  X 

O  Oi  O5  Oi  O5 


-*  -H  CO  -H  CO 

0  — t  CO  'N  t- 
•*  Oi  CO  GO  -N 
•CO  X  0  o:  p 

01  O5  O5  Ol  Oi 


CO  X  CO  Oi  -ti 
1—  ^H  CO  O  >O 

CO  CO  X  CO  00 

Oi  Oi  Oi  Oi  Oi 


-*  co  r—  i—  i— 

Oi  -t  Oi  -fi  Oi 

Si  TO  co  ~#  ~* 

CO  X  X  X  X 
Oi  Oi  Oi  Oi  Oi 


0  <o  co  co  t— 

CO  CO  CO  X  CO 

01  Oi  Oi  Oi  Oi 


GO  -H  -H  CO  C<l 
CO  CO  X  <N  t- 

•^  Oi  CO  X  "N 
I—  I—  CO  X  Oi 

GO   "/>  GO   CO  '-O 

Oi     Oi     Oi     Oi     Oi 


«O   -H   CO  -H 

CO  'N  I—  CO 

GO  co  r—  sq 


»O  O5  1<l  O  t—       Oi  — i  ^ 


CO  GO  X 


05  — '  CM  -N  ?1        ;M 

05  -fi  co  co  t-     sq 


<M  — i  O  CO 

"O  O  -* 
CO  — I  O 


CO  CO  X 
Oi  Oi  Oi 


O  »0  O       tO  — I  CO  — I  O       -H  CO  — t 
iC  O5  -t        X  CO  I—  (N  CO       rH  iO  O 


s^sii 

SoTSoT 


Oi  CO  CO  Oi  ^4 
iO  O  "^  Oi  CO 

0  -H    r*    rH    7<l          ^ 

X  X  X  X  X 

01  Oi  Oi  Oi  Oi 


CO  X 

Oi  Oi 


CO  CO  X  -*  Oi       CO  t-  f-i  -t  CO       ^  O  — i  'M  -M        'N  — i  O  Oi  1-       -t  — (  I h  Oi 

C>1  X  CO  Oi  — Ti       O  »O  ^H  CO  -H        ^  'M  1^-  'N  I"-       d  !•—  ^\  CO  ^H       CO  -H  iO  C-  -f 

OiOiOiOiOi       O5OiOiOiO5       OiOiOiOiOi       'OiOiOiOiOi       OiOiOiOiOi 


82i3£   §5^i^§  ^ii?S?2   Si^^v° 

02  co  ^  o 


^s?,1  s 
g£i    S' 


s  s  §  § 


O        ^    N    M  <*    >-5 
i-.      f.  i>.  r~  r^  i-. 

C5    CS    CS   CS   C5 


LOGARITHMS. 


I2S 


33333 


CO  CO  CM  TH 
Q  OS  CO   t—  i— l  »O  OS 
OS  CO  X   <M  t-  TH  »O 
CN  CN  CO  CO 


CO  >O  £"*  X  OS  OS  OS  X  £••  £O 

<N  O  O  H/i  X  <M  ^D  O  ^f  GO 

"     H/l  OS  CO  t-  <M  SO  •—  »O  OS 

?7  Cf  -*  "5h  »0  O  O 


-f  X  ^N  t-  TH 

c;  ^  t-  i—  oo 

os  os  os 

OS  OS  OS 


CO  OS  CO 

H/l  t-  l-H 


t—  OS  O  i— i 
t-  ^  ®  O 
OSCOXfMt— 
iH  94  C^  00  00 

OiCl 
OSOS 


x  01  -^  c; 

'Nt^^'^ 


§co  o  t-  -* 
H/I  x  TH  ira 
p  o  o  os 

i  OS  OS  OS       OS  Os  OS  OS  Os       OS'  OS  _.    _ . 
OSOSOSOSOS       OSOSOSOSOS       OSOSCSOS 


li 


X  —  •  O 
'NI^-^ 


>  >O  -f  CO       T-I  OS  t  --  f  O 


H^OSCOt-O       CO  "^  X  OS  O       —  — '  —  O  OS       1—  OCO<C 
OS  CO  X  01  1-       T-!  O  OS  CO  X        OJ  :C  O  ^f  I-       —  'J  x-  7 


2  £. 


O  o  os  co  o      os  i-i  -t  ic  o  t—  1^-  t—  :c  o  "t  *N  os  ^  co  cs  »r?  — '  >o 

O  OS  CO  X  JN       O  I-H  O  OS  CO       i iO  OS  CO  i f  X  -N  ^  C~.   "  'C 

"^*  X  CO  t-  S<l       tD  — i  O  OS  ^  X  CO  I-  ~*  ~  O  i.O  OS  CO  X  O^  O  TH  »O 

—       OS  OS  O  O  TH       TH  'N  Ol  Ol  CO  C?  -t  -t  o  £  O  ^  to  I-  t—  X  X  OS  OS 

iiiii  iiiis  iisii  iisi 


126 


HYPERBOLIC    LOGARITHMS. 


HYPERBOLIC   LOGARITHMS. 

The  hyperbolic  or  Napierian  logarithm  of  any  number  may 
be  obtained  by  multiplying  the  common  logarithm  by  the  con- 
stant 2.302585;  practically  2.3. 

Table  No.  6  gives  the  hyperbolic  logarithms  from  1.01  to 
30.  The  hyperbolic  logarithm  of  numbers  intermediate  between 
those  which  are  given  in  the  table  may  be  obtained  by  inter- 
polating proportional  differences. 


TADLE  No.  6.— Hyperbolic  or  Napierian  Logarithms 
of  Numbers. 


N 

Log. 

N 

Log. 

N 

Log. 

N 

Log. 

.01 

00099 

1.26 

02311 

1.51 

0.4121 

1.76 

0.5653 

.<>2 

0.0198 

1/27 

0/2-  ',90 

1.62 

0.4187 

1.77 

0.5710 

.03 

0.0  96 

1.28 

0/2469 

153 

0.4253 

1.78 

0.6766 

.04 

0  0:i92 

1.29 

0.2546 

1.54 

0.4318 

1.79 

0.5822 

.05 

00488 

1.30 

0.2624 

1  55 

0.4383 

1.80 

0.5878 

.06 

0.05S3 

1.31 

0  2700 

1.56 

0.4447 

1.81 

0.5933 

.07 

0.0677 

1.32 

0.2776 

157 

0.4511 

1.82 

06988 

.08 

00770 

1.33 

0.2S52 

1.58 

0.4574 

1.83 

0.6043 

.09 

00862 

1.31 

0.2927 

1.59 

0.4637 

1.84 

0.6098 

.10 

0.0953 

1.35 

03001 

1.60 

0.4700 

1.85 

0.6152 

.11 

0.1044 

1.36 

0.3075 

1.61 

0.4762 

1.86 

0.6206 

.12 

0.1133 

1.37 

O.:<148 

1.62 

0.4824 

1.87 

0.6259 

.13 

0.1222 

1.38 

0.3-J21 

1.63 

0.4886 

1.88 

0.6313 

.14 

01310 

1.39 

0  3-J'.i:» 

.64 

0.4947 

1.89 

0.6366 

.15 

0.1398 

1.40 

03365 

.65 

0.5008 

1.90 

0.6419 

.16 

0.1484 

1.41 

0.3436 

.66 

0.5068 

1.91 

0.6471 

.17 

0.1570 

1.42 

0.3507 

.67 

0.5128 

1.92 

0.6523 

.18 

0.  1  655 

1.43 

0.3577 

.68 

0.5188 

1.93 

0.6575 

.19 

0.1740 

1.44 

0.3646 

.69 

05247 

l.v»4 

0  H627 

1  .20 

0.1823 

1.45 

0.3716 

.70 

0.5306 

1.95 

0.6678 

.21 

0.1906 

1.46 

0.3784 

J.71 

0.5365 

1.96 

0.6729 

.22 

0.1988 

1.47 

0.3853 

1.72 

0.5423 

1.97 

0.6780 

.23 

0.2070 

1.48 

0.3920 

1.73 

0.5481 

1.98 

O.H831 

.24 

0.2151 

1.49 

0  3988 

1.74 

0.5539 

199 

O.H881 

1.25 

0.2231 

1.50 

0.4055 

1.75 

0.5596 

2.00 

0.6931 

HYPERBOLIC  LOGARITHMS. 


127 


N 

Log. 

N 

-  Log. 

N 

Log. 

N 

Log. 

•>.oi 

0.6981 

?.1l 

0.870ft 

2.S1 

1.0332 

3.21 

1.1063 

2.02 

0.7031 

2.42 

0.883S 

282 

10367 

3.22 

1.16-4 

2  03 

0.70SO 

2.43 

0>x;n 

2.83 

10403 

3.23 

1.1725 

2.04 

0.7129 

244 

0.89^0 

2.84 

1.0438 

3.24 

1.1756 

2.05 

0.7178 

2.45 

0.8961 

2.85 

1.0473 

3.25 

1.1787 

2.06 

0.7227 

2.46 

0.9002 

2.86 

1.0508 

3.28 

.1817 

2.07 

0.7275 

2.47 

0.9042 

2.87 

.0543 

3.27 

.1848 

2.08 

0.7324 

2.48 

0.90S3 

2.88 

.Oft78 

3/28 

.1878 

2.09 

0.7372 

2.49 

0.9123 

2.89 

.0613 

3.29 

.1909 

2.10 

0.7419 

2.50 

0.9163 

2.90 

.0647 

3.30 

.1939 

2.11 

0.7467 

2.51 

0.9203 

2.91 

.0682 

3.31 

.1969 

2.12 

0.7514 

H.62 

0.9243 

2.92 

.0716 

332 

.2000 

2.13 

0.7561 

2.53 

0.9282 

2.93 

0750 

3.33 

.2030 

2.14 

0.7608 

254 

0.93-J2 

2.94 

.0784 

3.34 

.2060 

2.15 

0.7655 

2.55 

09361 

2.95 

.0818 

3.35 

.2090 

2.16 

0.7701 

2.56 

0.91PO 

2.96 

.0852 

336 

.2119 

2.17 

0.7747 

2.57 

0.9439 

2.97 

.0886 

3.37 

.2149 

2.18 

07793 

2.58 

0.9478 

2.98 

.0919 

3.38 

1.2179 

2.19 

0.7839 

2.59 

0.9517 

2.99 

.0953 

3.39 

1.2208 

2.20 

0.7885 

260 

0.9555 

3. 

.0986 

3.40 

1.2238 

2.21 

0.7930 

261 

0.9594 

3.01 

.1019 

3.41 

1.2267 

2.22 

0.7975 

2-62 

0.9632 

3.02 

.1053 

3.42 

1.2296 

2.23 

08020 

2-63 

0.9670 

3.03 

.1086 

3.43 

1/23-26 

2.24 

0.8065 

2'64 

0.9708 

3.04 

1119 

3.44 

1.2355 

2.25 

0.8109 

265 

0.9746 

3.05 

.1151 

3.45 

1.2384 

2.26 

0.8154 

2.66 

0.9783 

3.06 

.1-184 

3.46 

1.2413 

2.27 

0.8198 

2.67 

098-21 

3.07 

.1217 

3.47 

1.2442 

2.28 

0.8242 

2.68 

0.98'.8 

3.08 

.l'_M9 

3.48 

1.2170 

2.29 

0.8286 

2.69 

0.9895 

3.09 

.U'S2 

3.49 

1.24'.»9 

2.30 

0.8329 

270 

0.9933 

3.10 

.1314 

3.50 

1.2528 

2.31 

0.8372 

2.71 

0.9969 

3.11 

.1346 

3.51 

1.2556 

2.32 

0.8416 

2.72 

1.0006 

3.12 

.1378 

3.52 

1.2:,85 

2.33 

0.8459 

2.73 

1.0043 

3.13 

.1410 

3.53 

1.2613 

2.34 

0.8502 

2.74 

1.0080 

3.14 

.1442 

3.54 

1.2641 

2.35 

0.8544 

2.75 

1.0116 

3.16 

.1474 

3.55 

1.2669 

2.36 

0.8687 

2.76 

1.0152 

3.16 

.1506 

3.56 

1.2H98 

2.37 

0.8629 

2.77 

1.0188 

3.17 

.1537 

3.57 

1.2726 

2.38 

0.8671 

2.78 

1.0225 

3.18 

.1569 

3.58 

1.2754 

2.39 

0.8713 

2.79 

1.0260 

3.19 

.1600 

3.59 

1.2782 

2.40 

0.8755 

2.80 

1.0296 

3.20 

.1632 

3.60 

1.2809 

128 


HYPERBOLIC  LOGARITHMS. 


N 

Log. 

N 

Log. 

N 

Log. 

N 

Log. 

3.61 

1.2837 

4.01 

1.3888 

4.41 

1.4839 

481 

1.5707 

362 

1.2865 

4.02 

1.3913 

4.42 

1.4801 

4.82 

1.5728 

3.63 

1  2892 

4.03 

1.3938 

4.43 

1.4884 

483 

1.5748 

364 

1.29'JO 

4.04 

1.3962 

4.44 

1.4907 

4.84 

1.5769 

3.65 

1.2947 

4.05 

1.3987 

4.45 

1.4929 

4.85 

1.5790 

3.06 

1.2976 

4.06 

1.4012 

4.46 

1.4951 

4.86 

1.5810 

3.67 

1.3002 

4.07 

1.4036 

4.47 

1.4974 

4.87 

1.5831 

3.68 

1.3029 

4.08 

1.4061 

4.48 

1.4996 

488 

1.5851 

3.69 

1.3056 

4.09 

1.4085 

4.49 

1.5019 

4.89 

1  5«72 

3.70 

1.3083 

4.10 

1.4110 

450 

1.5041 

4.90 

1.5892 

3.71 

1.3110 

4.11 

1.4134 

4.51 

1.5063 

4.91 

1.5913 

3.72 

1.3137 

4.12 

1.4159 

4.52 

1.5085 

492 

1.5933 

3.73 

1.3164 

4.13 

1.4183 

4.53 

1.5107 

4  93 

1.5953 

374 

1.3191 

414 

1.4207 

4.54 

1.5129 

4  94 

1.5974 

3.75 

1.3218 

4.15 

1.4231 

4.55 

1.5151 

4.95 

1.5994 

3.76 

1.32-14 

4.16 

1.4255 

4.56 

1.5173 

4  96 

1.6011 

3.77 

1.3271 

4.17 

1.4279 

4.57 

1.5195 

4.97 

1  GO;U 

3.78 

1.3297 

4.18 

1.4303 

4.58 

1.5217 

4.98 

1.6054 

3.79 

1.3324 

4.19 

1.4327 

4.59 

1.5239 

4.99 

1.6074 

3.80 

1.3350 

4.20 

1.4351 

4.60 

1.5261 

5. 

1.6094 

3.81 

1.3376 

4.21 

1.4375 

4.61 

1.5282 

6.01 

1.6114 

382 

1.3403 

4.22 

1.4398 

4.62 

1.5304 

5.02 

1.6134 

3.83 

1.3429 

423 

1.4422 

4.63 

1.5326 

5.03 

1.6154 

3.84 

1.3455 

4.24 

1.4446 

4.64 

1.6347 

5.04 

1.6174 

3.85 

1.3481 

4.25 

1.4469 

4.65 

1.5369 

5.05 

1.6194 

3.86 

1.3607 

4.26 

1.4493 

4.66 

1.5390 

5.06 

1.6214 

3.87 

1.3533 

4.27 

1.4516 

4.67 

1.5412 

5.07 

1.6233 

388 

1.3558 

4.28 

1.4540 

4.68 

1.5433 

5.08 

1.6253 

3.89 

1.3584 

4.29 

1.4563 

4.69 

1.5454 

5.09 

1.6273 

3.90 

1.3610 

4.30 

1.4586 

4.70 

1.5476 

5.10 

1.6292 

3.91 

1.3635 

431 

.4609 

4.71 

1.5497 

6.11 

1.6312 

3.92 

1.3661 

4.32 

.4633 

4.72 

1.5518 

5.12 

1.6332 

3.93 

1.3686 

4.33 

.4666 

473 

1  .5539 

5.13 

1.6351 

3.94 

1.3712 

4.34 

.4679 

4.74 

1.6.)60 

5.14 

16371 

3.95 

1.3737 

4.35 

.4702 

4.75 

1  5581 

5.16 

1.6390 

3.96 

1.3762 

4.36 

.4725 

4.76 

1.5602 

5.16 

1.6409 

397 

1.3788 

4.37 

.4748 

4.77 

1.5623 

5.17 

1.6429 

3.98 

1.3-13 

4.38 

.4770 

4.78 

1.56U 

5.18 

1.6448 

3.99 

1  3N38 

4.39 

.4793 

4.79 

1  r,6<i5 

5.19 

1.6467 

4. 

1.386s 

4.40 

.4816 

4.80 

1.6t)86 

5.20 

1.6487 

HYPERBOLIC   LOGARITHMS. 


129 


N 

Log. 

N 

Log. 

N 

Log. 

N 

Log. 

5.21 

16506 

6.61 

1.7246 

6.01 

1.7934 

6.41 

.8579 

5.22 

1.6525 

5.62 

1.7263 

6.02 

1.7951 

6.42 

.8594 

523 

1.6544 

5.03 

1.7281 

6.03 

1.7967 

6.43 

.8610 

5.24 

1.6563 

5.64 

1.7299 

6.04 

1.7984 

6.44 

.8625 

6.25 

1.6582 

5.65 

1.7317 

6.05 

1.8001 

6.45 

.86  i  I 

5.26 

1.6601 

5.66 

1.7334 

6.06 

1.8017 

646 

.8(556 

6.27 

1.6620 

6.67 

1.7352 

6.07 

1.8034 

6.47 

.8672 

5.28 

1.6639 

5.68 

1.7370 

6.08 

1.8050 

6.48 

.8687 

5.29 

1.6658 

5.69 

1.7387 

6.09 

1.8066 

6.49 

.8703 

6.30 

1.6677 

5.70 

1.7405 

6.10 

1.8083 

6.50 

.8718 

5.31 

1.6696 

5.71 

1.7422 

6.11 

1.8099 

6.51 

.8733 

5.32 

1.6716 

6.72 

1.7440 

6.12 

1.8116 

652 

.8749 

6.33 

1.6734 

5.73 

1  7457 

6.13 

1.8132 

6.53 

.8764 

6.34 

1.6752 

5.74 

1.7475 

6.14 

1.8148 

6.54 

.8779 

5.35 

1.6771 

6.76 

1.7492 

6.15 

1.8165 

6.55 

.8795 

5.36 

1.6790 

5.76 

1.7509 

616 

1.8181 

6.56 

8810 

6.37 

1.6808 

5.77 

1  7527 

6.17 

1.8197 

6.57 

.8H25 

6.38 

1.6827 

5.78 

1.7544 

6.18 

1.8213 

6.58 

18840 

6.39 

1.6845 

6.79 

1.7561 

6.19 

1.8229 

6.59 

1.8856 

5.40 

1.6864 

5.80 

1.7579 

6.20 

1.8245 

6.60 

1.8871 

541 

1.6882 

5.81 

1.7596 

6.21 

1.8262 

6.61 

1.8886 

6.42 

1.6901 

5.82 

1.7613 

6.22 

1.8278 

662 

1.8901 

5.43 

1.6919 

583 

1.7630 

6.23 

1.8294 

663 

.8916 

5.44 

1.6938 

5.84 

1.7647 

6.24 

1.8310 

6.64 

.8931 

6.45 

1.6956 

6.85 

1.7664 

6.25 

1.8326 

6.65 

.8946 

6.46 

1.6974 

5.86 

1.7681 

6.26 

1.8342 

6.66 

.8961 

6.47 

1.6993 

5.87 

1.7699 

6.27 

1.8358 

6.67 

.8976 

5.48 

1.7011 

5.88 

1.7716 

6.28 

18374 

6.68 

.8991 

549 

1.7029 

5.89 

1.7733 

6.29 

1.8390 

6.69 

.9006 

6.50 

1.7047 

5.90 

1.7750 

6.30 

1.8405 

6.70 

1.9021 

5.51 

1.70R6 

6.91 

1.7766 

6.31 

18421 

6.71 

1.9036 

5.52 

1.7084 

5.92 

1.7783 

6.32 

1.8437 

6.72 

1.9051 

6.53 

1.7102 

5.93 

1  7800 

6.33 

1.8-lo3 

6.73 

1.9066 

5.54 

1.71--'0 

5.94 

1.7817 

6.34 

1.8469 

6.74 

1.9081 

5.55 

1.7138 

5.95 

1.7834 

6.35 

1.848o 

6.76 

1.9095 

5.56 

1.7150 

6.96 

1.7851 

6.36 

1  .8500 

6.76 

1.9110 

6.57 

1.7174 

s  597 

1.7867 

6.37 

1.8516 

6.77 

1.9125 

5.68 

1.719-2 

6.98 

1.7884 

6.38 

1  8532 

6.78 

1.9140 

6.59 

1.7210 

5.99 

1  7901 

639 

1.8547 

6.79 

1.9155 

6.60 

1.7228 

6. 

1.7918 

6.40 

1.8663 

6.80 

1.9169 

130 


HYPERBOLIC   LOGARITHMS. 


N 

Log. 

N 

Log. 

N 

Log. 

N 

Log. 

6.81 

1.9184 

7.21 

1.9755 

7.61 

2.0205 

8.01 

2.08H7 

6.82 

1.9199 

7.22 

1.9769 

7.62 

2.0308 

802 

2.0819 

6.83 

1.9213 

7.23 

1.9782 

7.63 

2.0321 

8.03 

2.0s;?2 

6.84 

1.9228 

7.24 

1.9796 

7.64 

2.0334 

8.04 

2.  ('814 

6.85 

1.9242 

7.26 

1.9810 

7.65 

2.0347 

805 

2.0857 

6.86 

1.9257 

7.26 

1.9824 

7.66 

2.  0360 

8.06 

2.0869 

6.87 

1.9272 

7.27 

1.9838 

7.67 

2  0873 

8.07 

2.0882 

6.88 

1.9286 

7.28 

1.9851 

7.68 

2.0*86 

8.<8 

2.0894 

6.89 

1.9301 

7.29 

1.9865 

7.69 

2.0399 

8.09 

2.0906 

6.90 

1.9315 

7.30 

1.9879 

7.70 

2.0412 

810 

2.0919 

6.91 

1.9330 

7.31 

1.9892 

7.71 

2042^ 

8  11 

2.0931 

6.92 

1.9344 

7.32 

1.9906 

7.72 

2.0438 

8  12 

2.0943 

6.93 

1.9359 

7.33 

1.9920 

7.73 

2.0451 

8.13 

2.0956 

6.94 

1.9373 

7.34 

1.9933 

7.74 

2.0484 

8.14 

2.0968 

6.95 

1.9387 

7.36 

1.9947 

7.76 

2.0477 

8.15 

2.0980 

6.96 

1.9402 

7.36 

1.9961 

7.76 

2.0490 

8.16 

2.0992 

697 

1.9416 

7.37 

1.9974 

7.77 

2  0503 

8.17 

2.1005 

6.98 

1.9430 

7.38 

1.9988 

7.78 

2.0516 

8.18 

2.1017 

6.99 

1.9445 

7.39 

2.0001 

7.79 

2.0528 

8.19 

2.1029 

7. 

1.9459 

7.40 

2.0015 

7.80 

2.0541 

8.20 

2.1041 

7.01 

1.9473 

7.41 

2.0028 

7.81 

2.0554 

8.21 

21054 

7.02 

1.9488 

7.42 

2.0042 

7.82 

2.0567 

8.22 

2.1066 

7.03 

1.9502 

7.43 

2.0055 

7.83 

2.0580 

823 

2.1078 

7.04 

1.9516 

7.44 

2.0069 

7.84 

2.0592 

824 

2.1090 

7.05 

1.9530 

7.45 

2.0082 

7.85 

2.0605 

8.25 

2.1102 

7.06 

1.9544 

7.46 

2.0096 

7.86 

2.0618 

8.26 

2.1114 

7.07 

1.9559 

7.47 

2.0109 

7.87 

2.0631 

8.27 

2.1126 

7.08 

1.9573 

7.48 

2.0122 

7.88 

2.0M3 

8.28 

2.1138 

7.09 

1.9587 

7.49 

2.0136 

7.89 

2.0656 

8.29 

2.1150 

7.10 

1.9601 

7.50 

2.0149 

7.90 

2.0669 

8:30 

2.1163 

7.11 

1.9616 

7.51 

2.0162 

7.91 

2.068! 

831 

21175 

7.12 

1.9629 

7.52 

2.0176 

7.92 

2.0694 

8.32 

2.1187 

7.13 

1.9643 

7.53 

2.0189 

7.93 

2.0707 

8.33 

21199 

7.14 

1.9657 

7.54 

2.0202 

7.94 

2.0719 

8.34 

2.1211 

7.16 

1.9671 

7.55 

2.0215 

7.95 

2.0732 

8.35 

2.1223 

7.16 

1.9685 

7.56 

2.0229 

7.96 

2.0744 

8.36 

2.1235 

7.17 

1.9699 

7.57 

2.0242 

7.97 

2.0757 

8.37 

2.1247 

7.18 

1.9713 

7.58 

2.0255 

7.98 

2.0769 

8.38 

2.1258 

7.19 

1.9727 

7.59 

2.0268 

7.99 

2.0782 

8.39 

2.1270 

7.20 

1.9741 

7.60 

2.0281 

8. 

2.0794 

8.40 

2.1282 

HYPERBOLIC  LOGARITHMS. 


N 

Log. 

N 

j 

Log. 

N 

Log. 

N 

Log. 

8.41 

2.1294 

!  881 

2.1  7'.  9 

921 

2.2203 

9.61 

22628 

8.42 

2.13CK      8.82 

2.1770 

922 

2.2214 

9.62 

2.2638 

8.43 

2.1318 

883 

2.1782 

9.23 

22225 

9.63 

2.2649 

8.44 

2.1330 

8.84 

2.1793 

9.24 

2.2235 

9.64 

2.2659 

8.45 

2.1342 

8.85 

2.1804 

9.25 

2.2246 

9.65 

22670 

8.46 

2.1353 

8.86 

2.1815 

926 

2.2257 

9.66 

'  2  2680 

8.47 

2.13(>5 

8.87 

2.1827 

9.27 

2  2268 

9.67 

2.2690 

8.48 

2.1377 

8.88 

2.1838 

9.28 

2.2279 

9.68 

2.2701 

8.49 

2.1389 

8.89 

2.1849 

9.29 

2.2289 

9.69 

2.2711 

8.50 

2.1401 

8.90 

2.1861 

9.30 

2.2300 

9.70 

2.2721 

8.51 

2.1412 

8.91 

2.1872 

9.31 

2.2311 

9.71 

2.2732 

8.52 

2.1424 

892 

2.1883 

9.32 

2.2322 

9.72 

2.2742 

8.53 

2.1436 

8.93 

2.1894 

9.33 

2.2332 

9.73 

2.2752 

8.54 

2.1448 

8.94 

2.1905 

9.34 

2.2343 

9.74 

2.2762 

8.55 

2.1459 

8.95 

2.1917 

9.35 

2.2354 

9.75 

2.2773 

8.56 

2.1471 

8.96 

2.1928 

9.36 

2.2364 

9/76 

2.2783 

8.57 

2.1483 

8.97 

2.1939 

9.37 

2.2375 

9.77 

2.2793 

8.58 

2.1494 

8.98 

2.1950 

9.38 

2.2386 

9.78 

2.2803 

8.59 

2.1606 

8.99 

2.1961 

9.39 

2.2396 

9.79 

2.2814 

8.60 

2.1618 

9. 

2.1972 

9.40 

2.2407 

9.80 

2.2824 

8.61 

2.1629 

9.01 

2.1983 

9.41 

2.2418 

9.81 

2.2834 

862 

2.1541 

9.02 

2.1994 

9.42 

2.2428 

9.82 

2.2844 

8.63 

2.1552 

9.03 

2.2006 

9.43 

2.2439 

9.83 

2.2854 

8.64 

2.1564 

9.04 

2.2017 

9.44 

2.2450 

9.84 

2.2865 

8.65 

2.1576 

9.05 

2.2028 

9.45 

2.2460 

9.85 

2.2875 

8.66 

2.1587 

9.06 

2.2039 

9.46 

2.2471 

9.86 

2.2885 

8.67 

2.1599 

9.07 

2.2050 

9.47 

22481 

9.87 

2.2895 

8.68 

21610 

9.08 

2.2061 

9.48 

2.2492 

9.88 

2.2905 

8.69 

2.1622 

9.09 

2.2072 

9.49 

2.2^.02 

9.89 

2.2915 

8.70 

2.1633 

9.10 

2.2083 

9.50 

2.2513 

9.90 

2.2925 

8.71 

2.1645* 

9.11 

2.2094 

9.61 

2.2523 

9.91 

2.2935 

8.72 

2.1fio6 

9.12 

2.2105 

9.52 

2.2534 

9.92 

2.2946 

8.73 

2.1668 

9.13 

2.2116 

9.53 

2.2544 

9.93 

2.2956 

8.74 

2.1679 

9.14 

2.2127 

9.54 

22555 

9.94 

2.2966 

8.75 

2.1691 

9.16 

2.2138 

9.55 

2.2565 

9.95 

2.2976 

876 

2J702 

9.16 

2.2148 

9.66 

2.2576 

9.96 

22986 

8.77 

2.1713 

9.17 

2.2159 

9.67 

2.2686 

9.97 

2.2996 

8.78 

2.1725 

9.18 

2.2170 

9.58 

2.2597 

9.98 

2.3006 

8.79 

2.1736 

9.19 

2.2181 

9.59 

2.2007 

9.99 

2.3016 

8.80 

2.1748 

9.20 

2.2192 

9.60 

2.2618 

10. 

2.3026 

'32 


HYPERBOLIC  LOGARITHMS. 


N 

Log. 

N 

Log. 

N 

Log. 

N 

Log. 

10.1 

2.3126 

14.1 

2.6462 

18.1 

2.8959 

22.1 

3.0956 

10.2 

2.3225 

14.2 

2.6532 

18.2 

2.9014 

222 

3  1001 

10.3 

2.3322 

14.3 

2.6602 

183 

29069 

22.3 

31046 

10.4 

2.3419 

14.4 

2.6672 

18.4 

2.9123 

22.4 

3.1090 

10.5 

2.3515 

14.5 

2.6741 

18.5 

2.9178 

22.5 

3.1135 

10.6 

23609 

14.6 

2.6810 

18.6 

2  9231 

226 

3.1179 

10.7 

23703 

14.7 

2.6878 

18.7 

2  9285 

22.7 

3.1224 

10.8 

23796 

14.8 

2.6946 

18.8 

2.93;38 

22.8 

3.1267 

10.9 

23888 

14.9 

2.7013 

18.9 

2.9M1 

22.9 

3.1311 

11 

2.3979 

15. 

2.7080 

19. 

2.9444 

23. 

3.1355 

11.1 

2.4070 

15.1 

27147 

19.1 

2.9497 

23.1 

3.1398 

11.2 

2.4160 

15.2 

2.7213 

19.2 

2.9549 

232 

3.1441 

11.3 

2.4249 

15.3 

2.6279 

19.3 

2  9601 

23.3 

3  1484 

11.4 

2.4337 

15.4 

2.7344 

19.4 

2.9(553 

23.4 

31527 

11.6 

2.4424 

15.5 

2.7408 

19.5 

2.9704 

23.5 

3.1570 

11.6 

2.4510 

15.6 

2.7472 

19.6 

2.9755 

23.6 

3.1612 

11.7 

2461)6 

15.7 

2.7536 

197 

2.9806 

23.7 

3.1655 

11.8 

2  4681 

158 

2.7600 

198 

2.9856 

23.8 

3.1697 

11.9 

2.4765 

15.9 

2.7663 

19.9 

2.9907 

23.9 

3.1739 

12 

2.4849 

16. 

2.7726 

20. 

2.9957 

24. 

3.1780 

12.1 

2.4932 

16.1 

2.7788 

20.1 

30007 

24.1 

3.1822 

12.2 

2.5014 

16.2 

2.7850 

20.2 

3.0057 

24.2 

3.1863 

12.3 

2.5096 

16.3 

2.7912 

20.3 

3.0106 

24.3 

3.1905 

12.4 

2.5178 

16.4 

2.7973 

20.4 

3.0155 

24.4 

3  1946 

12.5 

2.5259 

16.5 

2.8033 

20.6 

3.0204 

24.5 

3.1987 

12.6 

2.6338 

16.6 

2.8094 

20.6 

3.0253 

24.6 

3.2027 

12.7 

25417 

16.7 

2.8154 

20.7 

3.0301 

24.7 

3.2068 

12.8 

2.5495 

16.8 

2.8214 

20.8 

3.0349 

24.8 

3.2108 

12.9 

2.5572 

16.9 

28273 

20.9 

3.0397 

24.9 

3.2149 

13 

2.5649 

17. 

2.8332 

21. 

3.0445 

25. 

3.2189 

13.1 

2.5726 

17.1 

2.8391 

21.1 

3.0493 

•25.5 

3.2387 

13.2 

2.5802 

17.2 

2.84  19 

21.2 

3.0540 

26. 

3.2581 

133 

2.5877 

17.3 

2.8507 

21.3 

3.0587 

26.5 

32771 

13.4 

2.5962 

17.4 

2.8565 

21.4 

3.0634 

27. 

3.2968 

13.6 

2.6027 

17.5 

2.8622 

21.5 

3.0680 

27.6 

3.3142 

13.6 

2.6101 

17.6 

2.8679 

21.6 

3.0727 

28. 

3.3322 

13.7 

2.6174 

17.7 

2.8735 

21.7 

3.0773 

28.6 

3.3499 

13.8 

2.6247 

17.8 

2.8792 

21.8 

3.0819 

29. 

33673 

13.9 

2.6319 

17.9 

28848 

21  9 

3.  OH  65 

29.5 

3.3844 

14. 

2.6391 

18. 

2.8904 

22. 

3.0910 

30. 

3.4012 

Weights  anb  Measures. 


The  yard  is  the  standard  unit  for  length  in  the  United 
States  and  Great  Britain.  To  determine  the  length  of  the  yard, 
a  pendulum  vibrating  seconds  in  a  vacuum  at  the  level  of  the 
sea  in  the  latitude  of  London,  with  the  Fahrenheit  thermometer 
at  62°,  is  supposed  to  be  divided  into  301,393  equal  parts  ;  360,- 
000  of  these  parts  is  the  length  of  the  standard  yard.  Actually, 
the  standard  yard  in  both  the  United  States  and  Great  Britain 
is  a  metallic  scale  made  with  great  care  and  kept  by  the  respec- 
tive governments,  and  from  this  standard  other  measures  of 
length  have  oeen  produced. 

The  standard  unit  of  weight  in  the  United  States  and 
Great  Britain  is  the  Troy  pound,  which  is  equal  in  weight  to 
22.2157  cubic  inches  of  distilled  water  at  62°  Fahrenheit,  the 
barometer  being  30  inches.  The  Troy  pound  contains  5,760 
Troy  grains;  the  Pound  Avoirdupois,  which  is  the  unit  of 
weight  used  in  commercial  transactions  and  mechanical  cal- 
culations in  the  United  States  and  Great  Britain,  is  equal  to 
7,000  Troy  grains. 

In  the  United  States  the  standard  unit  of  liquid  measure 
is  the  wine  gallon,  containing  231  cubic  inches  or  8.3389  pounds 
avoirdupois  of  distilled  water  at  a  temperature  of  its  greatest 
density  (39°-40°  F). 

In  the  United  States  the  standard  unit  for  dry  measure  is 
the  Winchester  Bushel,  containing  2150.42  cubic  inches. 

In  Great  Britain  the  standard  measure  for  both  liquid  and 
dry  substances  is  the  Imperial  Gallon,  which  is  defined  as  the 
volume  of  10  pounds  avoirdupois  of  distilled  water,  when  weighed 
at  62°  Fahrenheit  with  the  barometer  at  30  inches.  The  Im- 
perial Gallon  contains  277.463  cubic  inches.  The  Imperial 
Bushel  of  8  gallons  contains  2219.704  cubic  inches. 

Long  Measure. 

12  inches  =  1  foot  =  0.30479  meters. 

3  feet  =  1  yard  =  0.91437  meters. 

5>^  yards  =  1  rod  or  pole  =  16^  feet  =  198  inches. 

40  rods  =  1  furlong  =  220  yards  =  660  feet. 

8  furlongs  =  1  statute  or  land  mile  =  320  rods  =  1760  yards. 

3  miles  =  1  league  =  24  furlongs  =  960  rods. 

5280  feet  =  1  statute  or  land  mile  =  1.609  kilometer. 

1  geographical  or  nautical  mile  =  1  minute  =  ^  degree. 

(I33-) 


134  WEIGHTS   AND   MEASURES. 

As  adopted  by  the  British  admiralty,*  a  nautical  mile  is  6080  ft. 
I  nautical  mile  =  1.1515  statute  or  land  miles. 
1  statute  or  land  mile  =  0.869  nautical  miles. 

Square  Measure. 

1  square  yard  =  9  square  feet  =  0.836  square  meters. 
1  square  foot  =  144  square    inches  =  929  square  centi- 
meters. 

1  square  inch  =  6.4514  square  centimeters. 
A  section  of  land  is  1  mile  square  =  640  acres. 
1  acre  =  43,560  square  feet  =  0.40467  hectare. 
1  square  acre  is  208.71  feet  on  each  side. 

Cubic  fleasure. 

1  cubic  yard  =  27  cubic  feet  =  0.7645  cubic  meters. 

1  cubic  yard  =  201.97  (wine)  gallons  =  7.645  hectoliter. 

1  cubic  foot  =  1728  cubic  inches  =  28315.3  cubic  centi- 
meters. 

1  cubic  foot  ==  7.4805  (wine)  gallons  =  28.315  liters. 

NOTE. — 1  cubic  foot  contains  6.2355  imperial  (English)  gal- 
lons. 

A  cord  of  wood  =  128  cubic  feet,  being  4X4X8  feet. 

A  perch  of  stone  =  24^  cubic  feet,  being  IQ%  X  11A  X  1 
foot,  but  it  is  generally  taken  as  25  cubic  feet. 

Liquid  Hea&ure. 

1  pint  =  28.88  cubic  inches. 

2  pints  =  1  quart  =  57.75  cubic  inches  =  0.9463  liter. 
4  quarts  =  1  gallon  =  231  cubic  inches  =  3.7852  liters. 
NOTE. — 1  imperial  (English)  gallon  is  277.274  cubic  inches. 

Dry  Measure. 

1  standard  U.  S.  bushel  =  2150.42  cubic  inches. 

1  standard  U.  S.  bushel  =  4  pecks. 

1  peck  =  2  gallons  =  8  quarts. 

1  gallon  =  4  quarts  =  268*  cubic  inches. 

1  quart  =  2  pints  =  67|  cubic  inches. 

100  bushels  (approximately)  =  124>£  cubic  feet. 

80  bushels  (approximately)  =  100  cubic  feet. 

Avoirdupois  Weight. 

(Used  in  business  and  mechanical  calculations.) 

1  pound  =  16  ounces  =  0.45359  kilograms. 

1  ton  =  2240  pounds.     A  short  ton  is  2000  pounds. 

*  See  Machinery,  page  23,  Sept.,  1897. 


WEIGHTS    AND   MEASURES. 

Troy  Weight. 

(Used  when  weighing  gold,  silver  and  jewelry.) 

1  pound  =  12  ounces  =  0.37324  kilogram. 

1  ounce  =  20  pennyweights  =  1.0971  ounces  avdp. 

Apothecaries'  Weight. 

1  pound    =  1    pound  troy  weight  =  12  ounces. 
1  ounce     =  8    drachms. 
1  drachm  =  3  scruples. 
1  scruple  =  20  grains. 


135 


Weights  of  Produce. 

The  following  are  the  weights  of  certain  articles  of  produce: 


Pounds 
per  bushel. 

Wheat,                     60 
Corn  in  the  ear,      70 
Corn  shelled,          56 
Rye,                          56 
Buckwheat,            48 
Barley,                     48 

Pounds 
per  bushel. 

Oats,                     32 
Peas,                    60 
Ground  peas,      24 
Corn  meal,           48 
Malt,                    38 
White  beans,      60 

Pounds 
per  bushel. 

White  potatoes,     60 
Sweet  potatoes,     55 
Onions,                    57 
Turnips,                   57 
Clover  seed,            60 
Timothy  seed,        45 

The  fletric  System  of  Weights  and  Heasures. 

The  unit  in  the  metric  system  is  the  meter.     The  length  of 

(one  ten-millionth  part) 


1  ()  ooo  odd 


the  meter  was  intended  to  be  iv,uuuuuw  , 

of  the  length  of  a  quadrant  of  the  earth  through  Paris,  which 

is  the  same  as  TcrWoo<j<r  of  anY  quadrant  from  either  pole  to 

equator. 

By  later  calculations  it  has  been  ascertained  that  the  meter 
as  first  adopted  and  now  used  is  slightly  too  short  according  to 
this  theoretical  requirement,  but  this,  of  course,  makes  no  dif- 
ference ;  because,  practically  speaking,  the  length  of  a  meter  is 
the  length  of  a  certain  standard  meter  kept  at  Paris  in  the  care 
of  the  French  government,  and  it  is  from  this  standard  meter 
(and  not  from  the  quadrant  of  the  earth)  that  all  other  standard 
meters  kept  for  reference  are  derived. 

The  grain,  which  is  the  unit  of  weight,  is  the  weight  of  1 
cubic  centimeter  of  water  at  its  maximum  density,  which  is  at 
4°  C.  (39°  to  40°  F.) 

The  commercial  denomination  used  for  weight  is  the  kilo- 
gram =  1000  grams  =  1  cubic  decimeter  =  1  liter  of  water  at 
maximum  density. 

The  metric  system  has  been  adopted  by  Mexico,  Brazil, 
Chile  and  Peru,  and  by  all  European  countries  except  England 
and  Russia. 


136  WEIGHTS   AND   MEASURES. 

Length. 

1  Meter  =  10  Decimeters    =  39.37  inches. 

1  Decimeter    =  10  Centimeters  =  3.937  inches. 

1  Centimeter  =  10  Millimeters    =  0.3937  inches. 

1  Millimeter    =  ^v  Meter         =  0.03937  inches. 

1  Decameter  =  10  Meters  =  32  feet  9.7  inches. 

1  Hektometer  =  100  Meters         =328    "    1  inch. 

1  Kilometer     =  1000  Meters       =  0.6214  mile. 

1  Myriameter  =  10000  Meters     =  6.214  miles. 


Area. 

1  square  millimeter  =  0.00155  square  inch. 
100  square  millimeters  =  1  square  centimeter  =    0.155  sq.  inch. 
100      "      centimeters  =  1       "       decimeter   =15.5      sq.  inch. 
100      "      decimeters   =1       "       meter          =  10.764  sq.  feet. 

1  Centare  =  1      "       meter          =  1550  sq.  inches. 

1  Are  =100"       meters         =  119.6  sq.  yards. 

1  Hectare  =  10000  sq.  meters       =  2.471  acres. 


Solids. 

1  cubic  millimeter  =  0.000061  cubic  inch. 

1000  cubic  millimeters  =  1  cubic  centimeter  =  0.061  cubic  inch. 
1000      "     centimeters  =  1     "      decimeter  =  61.027     "        " 
1000      "     decimeters  =1    "      meter          =35.3        "     feet. 


Liquid. 

1  liter  =  10  deciliters  =  1      cubic  decimeter. 

1  deciliter    =  10  centiliters  =  100  cubic  centimeter. 

1  centiliter  =  10  milliliters  =  10    cubic  centimeters. 

1  milliliter   =  y^-  liters  =  1      cubic  centimeter. 

1  decaliter  =  10  liters  =  10    cubic  decimeters. 

1  hectoliter  =  100  liters  =  100  cubic  decimeters. 

1  kiloliter     =  1000  liters  =  1      cubic  meter. 
1  liter  =  61.027  cubic  inches  =  1.0567  quarts. 


WEIGHTS    AND    MEASURES.  137 

TABLE  No.  7.     Reducing  Millimeters  to  Inches. 


Mm. 

Inches. 

Mm. 

Inches. 

Mm. 

Inches. 

0.02 

0.00079 

0.52 

0.02047 

2 

0.07874 

0.04 

0.00157 

0.54 

0.02126 

3 

0.11811 

0.06 

0.00236 

0.56 

0.02205 

4 

0.15748 

0.08 

0.00315 

0.58 

0.02283 

5 

0.19685 

0.10 

0.00394 

0.60 

0.02362 

6 

0.23622 

0.12 

0.00472 

0.62 

0.02441 

7 

0.27559 

0.14 

0.00551 

0.64 

0.02520 

8 

0.31496 

0.16 

0.00630 

0.66 

0.02598 

9 

0.35433 

0.18 

0.00709 

0.68 

0.02677 

10 

0.39370 

0.20 

0.00787 

0.70 

0.02756 

11 

0.43307 

0.22 

0.00866 

0.72 

0.02835 

12 

0.47244 

0.24 

0.00945 

0.74 

0.02913 

13 

0.51181 

0.26 

0.01024 

0.76 

0.02992 

14 

0.55118 

0.28 

0.01102 

0.78 

0.03071 

15 

0.59055 

0.30 

0.01181 

0.80 

0.03150 

16 

0.62992 

0.32 

0.01260 

0.82 

0.03228 

17 

0.66929 

0.34 

0.01339 

0.84 

0.03307 

18 

0.70866 

0.36 

0.01417 

0.86 

0.03386 

19 

0.74803 

0.38 

0.01496 

0.88 

0.03465 

20 

0.78740 

0.40 

0.01575 

0.90 

0.03543 

21 

0.82677 

0.42 

0.01654 

0.92 

0.03622 

22 

0.86614 

0.44 

0.01732 

0.94 

0.03701 

23 

0.90551 

0.46 

0.01811 

0.96 

0.03780 

24 

0.94488 

0.48 

0.01890 

0.98 

0.03858 

25 

0.98425 

0.50 

0.01969 

1.00 

0.03937 

26 

1.02362 

TABLE  No.  8.    Reducing  Inches  to  Hillimeters. 

Inches.          Mm. 

Inches.        Mm. 

Inches.         Mm. 

TV                1-59 

\\  '        20.64 

2#           67.15 

y%          e.i7 

7/i           22.22 

2/2           63.50 

A                4.76 

if           23.81 

3               76.20 

X                6.35 

1                25.40 

4             101.6 

fV                7.94 

l*/&            28.57 

5             127 

y%          -9.52 

1#           31.75 

6             152.4 

TV                    11-11 

\Y%           34.92 

7            177.8 

Yz              12.70 

\l/2            38.10 

8             203.2 

14.29 

l>i           41.27 

9             228.6 

%              15.87 

1%            44.45 

10             254 

!  ,\               17.46 

17/S           47.62 

11              279.4 

%               19.05     |j        2                50.80             12              304.8 

138  WEIGHTS    AND    MEASURES. 

Table  of  Reduction  for  Pressure  per  Unit  of  Surface. 

1  kilogram  per  sq.  centimeter  =  14.223  pounds  per  sq.  inch. 
1  kilogram  per  sq.  centimeter  =    0.968  atmosphere. 
1  pound  per  sq.  inch  =  0.0703  kilograms  per  sq.  centimeter. 
1  pound  per  sq.  inch  =  0.068  atmosphere. 

Table  of  Reduction  for  Length  and  Weight. 

1  kilogram  per  kilometer  —  3.548  pounds  per  mile. 

1  kilogram  per  meter        =  0.672  pounds  per  foot. 

1  pound  per  mile  =  0.282  kilograms  per  kilometer. 

1  pound  per  foot  =  1.488  kilograms  per  meter. 

Weight  of  Water  (4°  C.) 

1  cubic  cm.  weighs  1  gram. 

1  cubic  inch  weighs  0.036125  pounds  =  16.386  grams. 
1  liter  weighs  1  kilogram  =  2.2046  pounds. 

1  quart  weighs  2.0862  pounds  —  0.9463  kilograms. 

1  cubic  meter  weighs  1000  kilograms  =  2204.6  pounds. 
1  cubic  foot  weighs  62.425  pounds      =  28.32    kilograms. 

Measure  of  Water. 

1  kilogram  measures  1  liter  =1.057   quarts. 

1  kilogram  measures  0.353  cubic  feet  =  61.03    cubic  inches. 

1  pound  measures       0.01602  cubic  ft.  =  0.454    liter. 

1  pound  measures      27.68  cubic  ins.  =  453.59  cubic  centimeters. 


SPECIFIC    GRAVITY. 

The  specific  gravity  of  a  body  is  its  weight  as  compared 
with  the  weight  of  an  equal  volume  of  another  body  which  ^  is 
adopted  as  a  standard.  For  all  solid  substances,  water  at  its 
maximum  density  (4°  C.)  is  the  usual  standard.  For  instance, 
the  specific  gravity  of  zinc  is  7 ;  this  simply  means  that  one 
cubic  foot  of  zinc  is  7  times  as  heavy  as  one  cubic  foot  of 
water.  One  cubic  foot  of  water  weighs  62.425  pounds.  There- 
fore, by  multiplying  the  specific  gravity  of  any  solid  body  by 
62.425  its  weight  per  cubic  foot  is  obtained.  In  the  metric 
system  of  measure  and  weight,  one  cubic  centimeter  of  water 
weighs  one  gram;  therefore  the  table  of  specific  gravity  will 
also  directly  give  the  weight  of  the  material  in  grams  per  cubic 
centimeter,  in  kilograms  per  cubic  decimeter,  or  in  1000  kilo- 
grams (the  so-called  metric  ton)  per  cubic  meter. 


WEIGHTS    AND    MEASURES.  139 

TABLE  No.  9.        Specific  Gravity,  Weights  and  Values. 


METALS. 


Metric. 


Kilog.  per 

;ubic  dec.  or 

specific 

gravity 


American. 


Pounds  per 
cubic  inch. 


Pounds  per 
cubic  foot. 


Approximate 

value 

per 

pound. 


Water  .  .  . 
Gold  (24  k)  . 
Platinum  .  . 
Silver  .  .  . 
Wrought  iron 
Cast  iron  .  . 
Tool  Steel  . 
Zinc  .... 
Antimony  .  . 
Copper  .  .  . 
Mercury  .  .  . 
Tin  .  .  . 
Aluminum  . 
Lead  . 


1 

19.361 
21.531 
10.474 

7.78 

7.21 

7.85 

7 

6.72 

8.607 

13.596 
7.291 
2.67 

11.36 


0.036125 

0.697 

0.775 

0.377 

0.28 

0.26 

0.284 

0.252 

0.242 

0.31 

0.489 

0.262 

0.096 

0.408 


62.425 
1208 
1344 

654 

485 

450 

490 

437 

419 

537 

849 

455 

166 

708 


$299.70 
122.00 
12.14 
0.015 
0.008 
0.10 
0.10 
0.12 
0.15 

0.25 
0.05 


TABLE  No.  10.    Specific  Gravity  and  Weight  of  Medium 
Dry   Wood. 


VARIETY. 


Metric. 


Kilog.  per  cubic 
dec.  specific  gravity. 


American. 


Pounds   per    cubic 
foot. 


Birch 

Ash 

Beech  

Oak 

Ebony 

Lignum-vitas      .    . 
Spanish  mahogany 

Hickory 

Spruce 

Pine 

Pitch  pine  .... 


0.60  to  0.80 

0.50  to  0.80 

0.60  to  0.80 

0.60  to  0.90 

1.19 

1.33 

0.85 

0.50 

0.50 

0.40  to  0.80 
0.80 


37.5  to  50 

31     to  50 

37.5  to  50 

37.5  to  56 

74 

83 

53 

32 

32 

25  to  50 
50 


140 


WEIGHTS    AND    MEASURES. 


TABLE  No.  ii.    Specific   Gravity  and  Weight  per  Cubic 
Foot  of  Various  Haterials. 

(The  weight  may  vary  according  to  the  properties  of  the  material). 


MATERIALS. 


Metric. 


Kilog.  per  cubic  dec. 
specific  gravity. 


American. 


Pounds 
per  cubic  foot. 


Asphalt  .... 

Brick 

Gray  granite  .  . 
Red  granite  .  . 
Limestone  .  .  . 

Sand 

Portland  cement 
Brickwork  .  . 

Slate    

Glass 

Emery  .... 
Grindstone  .  . 

Coal 

Porcelain  .  .  . 
Lime  . 


1.4 

1.6  to  2 

2.4 

2.5  to  3 

2.7 

1.5 

1.26 

1.75 

2.8 

2.52 

4.0 

2.4 

1.5 

2.4 

0.96 


87 
100  to  125 

150 

157  to  187 
168 

94 

78 
110 
175 
157 
250 
150 

64 
150 

60 


TABLE  No.  12.    Specific  Gravity  and  Weight  of  Liquids. 


LIQUIDS. 


0    > 


Metric. 


Kilog. 

per  cubic 

dec. 


Kilog. 
per 
liter. 


American. 


Pounds 

per 
cub.  inch. 


Pounds 
gallon. 


Water     .   .   . 
Sea  water     . 
Sulphuric  acid 
Muriatic  acid 
Nitric  acid    . 
Alcohol  .    .    . 
Linseed  oil    . 
Turpentine    . 
Petroleum 
Machine  oil  . 


1 

1.03 

1.841 

1.2 

1.217 

0.833 

0.94 

0.87 

0.878 

0.9 


1 

1.03 

1.841 

1.2 

1.217 

0.833 

0.94 

0.87 

0.878 

0.9 


.03 

.841 

.2 

.217 
0.833 
0.94 
0.87 
0.878 
0.9 


0.036125 

0.037 

0.067 

0.043 

0.044 

0.03 

0.034 

0.031 

0.032 

0.0324 


8.33 
8.55 

15.48 
9.93 

10.16 
6.93 
7.85 
7.16 
7.39 
7.5 


WEIGHTS    AND    MEASURES.  141 

To  Calculate  Weight  of  Casting  from  Weight  of  Pattern. 

When  pattern  is  made  from  pine  and  no  nails  used,  .the 
rule  is  :  Multiply  the  weight  of  the  pattern  by  17  and  the  prod- 
uct is  the  weight  of  the  castings. 

When  nails  are  used  in  the  pattern,  multiply  its  weight  by 
a  little  less,  probably  15  or  16. 

When  the  pattern  has  core  prints,  their  weight  must  be 
calculated  and  also  the  weight  of  what  there  is  to  be  cored  out 
in  the  casting,  which  must  all  be  deducted.  This  mode  of  cal- 
culating the  weight  of  castings  is,  of  course,  only  approxima- 
tion, but  it  is  frequently  very  useful. 


Weight  of  an  Iron  Bar  of  any  Shape  of  Cross  Section. 

A  wrought  iron  bar  of  1  square  inch  area  of  cross  section 
and  one  yard  long  weighs  10  pounds.  Therefore,  the  weight  of 
wrought  iron  bars  of  any  shape,  as,  for  instance,  railroad  rails, 
I  beams,  etc.,  may  very  conveniently  be  obtained  by  first  mak- 
ing a  correct,  full  size  drawing  of  the  cross  section  and  measur- 
ing its  area  by  a  planimeter,  which  gives  the  area  in  square 
inches.  Multiply  this  area  by  10  and  the  product  is  the  weight 
in  pounds  per  yard;  or  multiply  the  area  by  3.33  and  the  product 
is  the  weight  in  pounds  per  foot. 


To  Calculate  Weight  of  Sheet  Iron  of  any  Thickness. 

One  square  foot  of  wrought  iron,  1  inch  thick,  weighs  very 
nearly  40  pounds  (40.2  pounds)  and  one  square  foot  ¥y,  which 
is  joffo  thick,  weighs  1  pound.  Therefore,  a  practical  rule  for 
quick  calculation  of  the  weight  of  sheet  iron  is:  Divide  the 
thickness  of  the  iron  as  measured  by  a  micrometer  calliper  in 
thousandths  of  inches  by  25,  and  the  quotient  is  the  weight  in 
pounds  per  square  foot. 


To  Calculate  the  Weight  of  Metals  Not  Given  in  the  Tables. 

Find  the  weight  of  wrought  iron,  and  multiply  by  the  fol- 
lowing constants : 

Weight  of  wrought  iron  X  0.928  =  cast  iron. 

"  X  1.014  =  steel. 

"         "          "          "  X  0.918  =  zinc. 

"         "          "          "  X  1.144  =  copper. 

«'         «          «          «'  X  1.468  ~  lead, 


142 


WEIGHTS   AND    MEASURES. 


To  Calculate  the  Weight  of  Zinc,  Copper,  Lead,  etc., 
in  Sheets. 

First  find  the  weight  by  the  rule  given  for  sheet  iron,  and 
multiply  by  the  constant  as  given  in  the  above  table,  and  the 
product  is  the  weight  of  each  metal  in  pounds  per  square  foot. 

To  Calculate  the  Weight  of  Cast  Iron  Balls. 

Multiply  the  cube  of  the  diameter  in  inches  by  0.1377,  and 
the  product  is  the  weight  of  the  ball  in  pounds. 
Thus  : 


W—D^y.  0.1377.  D  =  1. 

D  =  diameter  of  ball  in  inches. 
W  =  weight  of  ball  in  pounds. 

In  metric  measure,  multiply  the  cube  of  the  diameter  in 
centimeters  by  0.003775,  and  the  product  is  the  weight  of  the 
ball  in  kilograms. 
Thus  : 

IV  =  M*  X  0.003775.       M  —  6.422  X 

W  =  weight  in  kilograms. 

M  —  diameter  of  ball  in  centimeters. 


TABLE  No.  13.    Weight  of  Round  Steel  per  Lineal  Foot. 

Steel  weighing  489  pounds  per  Cubic  Foot. 


Diameter   in 
inches. 

Weight 
Per  Foot. 

Diameter  in 
Inches. 

Weight  Per 
Foot. 

Diameter  in 
Inches. 

Weight  Per 
Foot. 

A 

.0104 

1    TV 

3.011 

2    */s 

12.044 

.042 

y& 

3.375 

i/ 

13.503 

3_ 

.094 

T36 

3.761 

N 

15.045 

1A 

.167 

X 

4.168 

16.67 

A 

.261 

A 

4.595 

X 

18.379 

.375 

&& 

5.043 

20.171 

A 

.511 

Tff 

5.512 

H 

22.047 

\/ 

.667 

H 

6.001 

3 

24.005 

T9* 

.844 

9 

6.512 

# 

26.048 

1.042 

ff 

7.043 

X 

28.173 

8 

1.261 

7.596 

30.382 

¥ 

1.5 

X 

8.169 

\/2 

32.674 

it 

1.761 

8.702 

ft 

35.05 

2.042 

7A 

9.377 

% 

37.508 

if 

2.344 

it 

10.013 

40.05 

l 

2.667 

2 

10.669 

4 

42.675 

WEIGHTS   AND    MEASURES. 


TABLE  No.  1 4.     Weights  of  Square  and  Round  Bars  of 
Wrought  Iron  in  Pounds  per  Lineal  Foot. 

(Iron  weighing  480  pounds  per  cvibic  foot). 


Thickness 
or  Dinmeter 
in 
Inches. 

Weight  of 
Square  Bar 
One  Foot 
Long. 

Weight  of 
Round  Bar 
One  Foot 
Long, 

Thickness 
or   Diameter 
in 
Inches. 

Weight  of 
Square  Bar 
One  Foot 
Long. 

Weight  of 
Round   Bar 
One  Foot 
Long. 

VlO 

.013 

.010 

2       %e 

21.89 

17.19 

% 

.052 

.041 

% 

22.97 

18.04 

%e 

.117 

.092 

H16 

24.08 

18.91 

% 

.208 

.164 

% 

25.21 

19.80 

r/io 

.320 

.256 

13/io 

26.37 

20.71 

% 

.469 

.368 

% 

27.55 

21.64 

7/10 

.638 

.501 

"Ke 

28.76 

22.59 

X 

.833 

.654 

3 

30 

23.56 

°/16 

1.055 

.828 

Vie 

31.26 

24.55 

% 

1.302 

1.023 

% 

32.55 

25.57 

U/10 

1.576 

1.237 

3/16 

33.8-7 

26.  60 

% 

1.875 

1.473 

% 

35.21 

27.65 

I:HG 

2.201 

1.728 

5/16 

36.58 

28.73 

% 

2.551 

2.004 

% 

37.97 

29.82 

lr/io 

2.930 

2.301 

7/16 

39.39 

30.94 

1 

3.333 

2.618 

I/ 
/2 

40.83 

32.07 

VlO 

3.763 

2.955 

%e 

42.30 

33.23 

% 

4.219 

3.313 

% 

43.80 

34.40 

•Yio 

4.701 

3.692 

ll/16 

45.33 

35.60 

% 

5.208 

4.091 

% 

46.88 

36.82 

%e 

5.742 

4.510 

13/16 

48.45 

38.05 

% 

6.302 

4.1)50 

% 

50.05 

39.31 

7/10 

6.888 

5.410 

lr/16 

51.68 

40.59 

% 

7.5 

5.890 

4 

53.33 

41.89 

%0 

8.138 

6.392 

Mo 

55.01 

43.21 

% 

8.802 

6.913 

% 

56.72 

44.55 

Hio 

9.492 

7.455 

*/ic 

58.45 

45.91 

% 

10.21 

8.018 

% 

60.21 

47.29 

i-Tio 

10.95 

8.601 

%e 

61.99 

48.69 

% 

11.72 

9.204 

% 

(53.80 

50.11 

lr/io 

12.51 

9.828 

He 

65.64 

51.55 

2 

13.33 

10.47 

X 

67.50 

53.01 

He 

14.18 

11.14 

%G 

09.39 

54.50 

% 

15.05 

11.82 

% 

71.30 

56 

3/10 

15.95 

12.53 

Vl6 

73.24 

57.52 

% 

16.88 

13.25 

% 

75.21 

59.07 

5/16 

17.88 

14 

13/10 

77.20 

60.63 

% 

18.80 

14.77 

% 

79.22 

62.22 

!/io 

19.80 

15.55 

ir/io 

81.26 

63.82 

X 

20.83 

16.36 

5 

83.33 

65.45 

144 


WEIGHTS    AND    MEASURES. 


TABLE    No.   i4.  — (Continued). 


Thickness 
or  Diameter 
in 

Weight  of 
Square  Bar 
One  Foot 

Weight  of 
Round  Bar 
One  Foot 

Thickness 
or  Diameter 
in 

Weight  of 
Square  Bar 
One  Foot 

Weight  of 
Round  Bar 
One  Foot 

Inches. 

Long. 

Long. 

Inches. 

Long. 

Long. 

5       I/LQ 

85.43 

67.10 

7  \ 

169.2 

132.9 

X 

87.55 

68.76 

175.2 

137.6 

3/16 

89.70 

70.45 

% 

181.3 

142.4 

X 

91.88 

72.16 

5    ^ 

187.5 

147.3 

5/16 

94.08 

73.89 

193.8 

152.2 

X 

96.30 

75.64 

3^ 

200.2 

157.2 

%6 

98.55 

77.40 

% 

206.7 

162.4 

X 

100.8 

79.19 

8 

213.3 

167.6 

%6 

103.1 

81.00 

X 

226.9 

178.2 

X 

105.5 

82.83 

X 

240.8 

189.2 

ll^Q 

107.8 

84.69 

255.2 

200.4 

% 

110.2 

86.56 

9  4 

270.0 

212.1 

13/16 

112.6 

88.45 

i/ 

285.2 

224.0 

X 

115.1 

90.36 

i/ 

300.8 

236.3 

15/16 

117.5 

92.29 

% 

316.9 

248.9 

6 

120.0 

94.25 

10 

333.3 

261.8 

X 

125.1 

98.22 

3£ 

350.2 

275.1 

X 

130.2 

192.3 

X 

367.5 

288.6 

% 

135.5 

106.4 

% 

385.2 

302.5 

X 

140.8 

110.6 

11 

403.3 

316.8 

X 

146.3 

114.9 

i/ 

421.9 

331.3 

% 

151.9 

119.3 

i/ 

440.8 

346.2 

X 

157.6 

123.7 

3X 

460.2 

361.4 

7 

163.3 

128.3 

12 

480. 

377. 

TABLE  No.  15.    Weight  of  Flat  Iron  in  Pounds  per  Foot. 


Inches 

Vl6 

% 

3/16 

X 

5/16 

% 

7/10 

X 

% 

% 

% 

1 

X 

0.11 

0.21 

0.32 

0.42 

0.53 

0.63 

0.73 

0.84 

% 

0.13 

0.26 

0.40 

0.53 

0.66 

0.79 

0.92 

1.06 

1.31 

% 

0.16 

0.32 

0.47 

0.63 

0.79 

0.95 

1.11 

1.26 

1.58 

1.90 

% 

0.18 

0.37 

0.55 

0.74 

0.92 

1.11 

1.29!  1.48 

1.85 

2.22 

2.58 

0.21 

0.42 

0.03 

0.84 

.05 

1.26 

1.47 

1.68 

2.11 

2.53 

2.95 

3.37 

IX 

0.24 

0.47 

0.71 

0.95 

.18 

1.42 

1.66 

1.90 

2.37 

2.84 

3.32;  3  79 

IX 

0.26 

0.53 

0.79 

1.05 

.32 

1.58 

1.84 

2.11 

2.63 

3.16 

3.68 

4.21 

1% 

0.29 

0.58 

0.87 

1.16 

.45 

1.74!2.03 

2.32 

2.89 

3.47 

4.05 

463 

IX 

0.32 

0.63 

0.95 

1.26 

.58 

1.90 

2.21 

2.53 

3.16 

3.79 

4.42 

5.05 

IX 

0.34 

0.68 

1.03 

1.37 

.71 

2.05 

2.39j  2.74 

3.42 

4.11 

4.79 

5.47 

1% 

0.37 

0.74 

1.11 

1.47 

1.84 

2.21 

2.58 

2  95  3.68 

4.42 

5.16J  5.89 

1% 

0.40 

0.79 

1.18 

1.58 

1.97 

2.37 

2.76 

3.16 

3.95 

4.74 

5.53 

6.32 

2 

0.42 

0.84 

1.26 

1.68 

2.11 

2.53  2.95 

3.37 

4.21 

5.05 

5.89 

6.74 

TABLE  No.  16.    Sizes  of  Numbers  of  the  U.  S.  Standard 
Gage  for  Sheet  and  Plate  Iron  and  Steel. 

(Brown  &  Sharpe  Mfg.  Co.) 


Number 
of 
Gage. 

Approximate 
Thickness  in 
Fractions  of 
an  Inch. 

Approximate 
Thickness  in 
Decimal  Parts 
of  an  Inch. 

Weight  Per 
Square  Foot 
in  Ounces 
Avoirdupois. 

Weight  Per 
Square  Foot 
in  Pounds 
Avoirdupois. 

0000000 

X 

.5 

320 

20.00 

000000 

If 

.46875 

300 

18.75 

00000 
0000 

1 

.4375 
.40625 

280 
260 

17.50 
16.25 

000 

H 

.375 

240 

15. 

00 

H 

.34375 

220 

13.75 

0 

* 

.3125 

200 

12.50 

1 

.28125 

180 

11.25 

2 

IT 

.265625 

170 

10.625 

3 

*% 

.25 

160 

10. 

4 

8 

.234375 

150 

9.375 

5 

A 

.21875 

140 

8.75 

6 

If 

.203124 

130 

8.125 

7 
8 

I 

.1875 
.171875 

120 
110 

7.5 

6.875 

9 

A 

.15625 

100 

6.25 

10 

& 

.140625 

90 

5.625 

11 

% 

.125 

80 

5. 

12 

A 

.109375 

70 

4.375 

13 

A 

.09375 

60 

3.75 

14 

A 

.078125 

50 

3.125 

15 
16 

f 

.0703125 
.0625 

45 
40 

2.8125 
2.5 

17 

* 

.05625 

36 

2.25 

18 

A 

.05 

32 

2. 

19 

T 
T6<* 

.04375* 

28 

1.75 

20 

83* 

.0375 

24 

1.50 

21 

aVcr 

.034375 

22 

1.375 

22 

* 

.03125 

20 

1.25 

23 

71T5" 

.028125 

18 

1.125 

24 

.025 

16 

1. 

25 

~5%~5 

.021875 

14 

.875 

26 

T6T7 

.01875 

12 

.75 

27 

28 

f 

.0171875 
.015625 

11 

10 

.6875 
.625 

29 

9 
tf¥tf 

.0140625 

9 

.5625 

30 

1 
"8% 

.0125 

8 

.5 

31 

<54]) 

.0109375 

7 

.4375 

32 

tflb 

.01015625 

W 

.40625 

33 

db 

.009375 

6 

.375 

34 

5h 

.00859375 

5^ 

.34375 

35 

db 

.0078125 

5 

.3125 

36 

9 

T2TTF 

.00703125 

4K 

.28125 

37 

rttjf 

.006640625 

4# 

.265625 

38 

db 

.00625 

4 

.25 

(145) 


146 


WEIGHTS    AND    MEASURES. 


TABLE  No.  1 7.      Different  Standards  for  Wire  Gage  in 
Use  in  the  United  States. 

Dimensions  of  Sizes  in  Decimal  Parts  of  an  Inch. 
(Brown  &  Sharpe  Mfg.  Co.) 


Number  of 
Wire  Gage. 

American 
or  Brown  & 
Sharpe. 

Birmingham 
or 
Stubs'  Wire. 

Washburn  & 
Moen  Mfg.  Co. 
Worcester, 

Mass. 

Trenton  Iron 
Co.,  Trenton, 
N.J. 

g 

|S 

wg 

£ 

-d  jj 

££ 
tfl'jj 
E> 

Number  of 
Wire  Gage. 

000000 

•46875 

000000 

00000 

.45 

•4375 

00000 

0000 

.46 

.454 

.3938 

.4 

•40625 

0000 

000 

.40964 

.425 

.3625 

.36 

.375 

000 

00 

.3648 

.38 

.3310 

.33 

•34375 

00 

0 

.32486 

.34 

.3065 

.305 

.3125 

0 

1 

.2893 

.3 

.2830 

.285  ' 

.227 

.28125 

1 

2 

.25763 

.284 

.2625 

.265 

.219 

.265625 

2 

3 

.22942 

.259 

.2437 

.245 

.212 

.25 

3 

4 

.20431 

.238 

.2253 

.225 

.207 

.234375 

4 

5 

.18194 

.22 

.2070 

.205 

.204 

.21875 

5 

G 

.16202 

.203 

.1920 

.19 

.201 

.203125 

6 

7 

.14428 

.18 

.1770 

.175 

.199 

.1875 

7 

8 

.12849 

.165 

.1620 

.16 

.197 

.171875 

8 

9 

.11443 

.148 

.1483 

.145 

.194 

.15625 

9 

10 

.10189 

.134 

.1350 

.13 

.191 

.140625 

10 

11 

.090742 

.12 

.1205 

.1175 

.188 

.125 

11 

12 

-080808 

.109 

.1055 

.105 

.185 

.109375 

12 

18 

.071961 

.095 

.0915 

.0925 

.182 

.09375 

13 

14 

.064084 

.083 

.0800 

.08 

.180 

.078125 

14 

15 

.057068 

.072 

.0720 

.07 

.178 

.0703125 

15 

10 

.05082 

.065 

.0625 

.061 

.175 

.0625 

16 

17 

.045257 

.058 

.0540 

.0525 

.172 

.05625 

17 

18 

.040303 

.049 

.0475 

.045 

.168 

.05 

18 

10 

.03589 

.042 

.0410 

.04 

.164 

.04375 

19 

20 

.031961 

.035 

.0348 

.035 

.101 

.0375 

20 

21 

.028462 

.032 

.03175 

.031 

.157 

.034375 

21 

22 

.025347 

.028 

.0286 

.028 

.155 

.03125 

22 

23 

.022571 

.025 

.0258 

.025 

.153 

.028125 

23 

24 

.0201 

.022 

.0230 

.0225 

.151 

.025 

24 

25 

.0179 

.02 

.0204 

.02 

.148 

.021875 

25 

20 

.01594 

.018 

.0181 

.018 

.146 

.01875 

26 

27 

.014195 

.016 

.0173 

.017 

.143 

.0171875 

27 

28 

.012641 

.014 

.0162 

.016 

.139 

.015625 

28 

20 

.011257 

.013 

.0150 

.015 

.134 

.0140625 

29 

30 

.010025 

.012 

.0140 

.014 

.127 

.0125 

30 

WEIGHTS  AND    MEASURES. 


TABLE  No.  17.  — (Continued). 


jiJ 

111 

1  -s 

111  3 

11 

1—1  c  • 

„  jj 

fjj 

^S 

11 

u 

g£j= 

l°i 

•s  e  f  3 

!*"_£ 

c/}"w 

|l 

** 

0 

a  1 

£jp 

£6 

21 

C/3 

t> 

31 

.008928 

.01 

0132 

.013 

.120 

.0109375 

31 

32 

.00795 

.009 

.0128 

.012 

.115 

.01015625 

32 

33 

.00708 

.008 

.0118 

.011 

.112 

.009375 

33 

34 

.006304 

.007 

.0104 

.01 

.110 

.00859375 

34 

35 

.005614 

.005 

.0095 

.0095 

.108 

.0078125 

35 

36 

.005 

.004 

.0090 

.009 

.106 

.00703125 

36 

37 

.004453 

.0085 

.103 

.006640625 

37 

38 

.003965 

.008 

.101 

.00625 

38 

39 

.003531 

.0075 

.099 

39 

40 

.003144 

.007 

.097 

40 

TABLE  No.  1 8.  Weight  of  Iron  Wire  in  Pounds  per  1 00  Feet. 


No.  of 
Wire 
Gage. 

American 
or  Brown  & 
Sharpe. 

Birmingham 
or 
Stubs'  Wire. 

No.  of 
Wire 
Gage. 

American 
or  Brown  & 
Sharpe. 

Birmingham 
or 
Stubs'  Wire. 

0000 

56.074 

54.620 

19 

0.341 

0.467 

000 

44.4683 

47.865 

20 

0.270 

0.324 

00 

35.265 

38.200 

21 

0.214 

0.271 

0 

27.9(5(5 

30.634 

22 

0.170 

0.207 

1 

22.178 

23.850 

23 

0.135 

0.165 

2 

17.588 

21.373 

24 

0.107 

0.128 

3 

13.948 

17.776 

25 

0.0849 

0.106 

4 

11.061 

15.010 

26 

0.0673 

0.0858 

5 

8.772 

12.826 

27 

0.0534 

0.0678 

6 

6.956 

10.920 

28 

0.0423 

0.0519 

7 

5.516 

8.586 

29 

0.0335 

0.0447 

8 

4.375 

7.214 

30 

0.0266 

0.0381 

9 

3.469 

5.804 

31 

0.0211 

0.0265 

10 

2.751 

4.758 

32 

0.0167 

0.0214 

11 

2.182 

3.816 

33 

0.0132 

0.0169 

12 

1.730 

3.148 

34 

0.0105 

0.0129 

13 

1.372 

2.391 

35 

0.00836 

0.00662 

14 

1.088 

1.825 

36 

0.006(52 

0.00424 

15 

0.868 

1.372 

37 

0.00525 

16 

0.684 

1.119 

38 

0.00416 

17 

0.542 

0.891 

39 

0.00330 

18 

0.430 

0.636 

40 

0.00262 

TABLE   No.  10.— Decimal  Equivalents  of   the  Numbers  of   Twist  Drill 
and  Steel  Wire  Gage.     (Brown  &  Sharpe  Mfg.  Co.) 


Size  in 

Size  in 

Size  in 

Size 

Size 

Size  in 

No. 

Deci- 

No. 

Deci- 

No. 

Deci- 

No. 

in 

No. 

in 

No. 

Deci- 

mals. 

mals. 

mals. 

Deci- 
mals. 

Deci- 
mals. 

mals. 

1 

.2280 

15 

.1800 

29 

.1360 

42 

.0935 

55 

.0520 

68 

.0310 

2 

.2210 

16 

.1770 

30 

.1285 

43 

.0890 

56 

.0465 

69 

.02925 

3 

.2130 

17 

.1730 

31 

.1200 

44 

.0860 

57 

.0430 

70 

.0280 

4 

.2090 

18 

.1695 

32 

.1160 

45 

.0820 

58 

.0420 

71 

.0260 

5 

.2055 

19 

.1660 

33 

.1130 

46 

.0810 

59 

.0410 

72 

.0250 

6 

.2040 

20 

.1610 

34 

.1110 

47 

.0785 

60 

.0400 

73 

.0240 

7 

.2010 

21 

.1590 

35 

.1100 

48 

.0760 

61 

.0390 

74 

.0225 

8 

.1990 

22 

.1570 

36 

.1065 

49 

.0730 

62 

.0380 

75 

.0210 

9 

.1960 

23 

.1540 

37 

.1040 

50 

.0700 

63 

.0370 

76 

.0200 

10 

.1935 

24 

.1520 

38 

.1015 

51 

.0670 

64 

.0360 

77 

.0180 

11 

.1910 

25 

.1495 

39 

.0995 

52 

.0635 

65 

.0350 

78 

.0160 

12 

.1890 

26 

.1470 

40 

.0980 

53 

.0595! 

0(5 

.0330 

79 

.0145 

13 

.1850 

27 

.1440 

41 

.0960 

54 

.0550 

67 

.0320 

80 

.0135 

14 

.1820 

28 

.1405 

1 

TABLE  No.  20.—  Decimal  Equivalents  of  Stubs'  Steel  Wire  Gage. 

(  Brown  &  Sharpe  Mfg.  Co.) 

. 

Size 

•o« 

Size 

-I 

II      8> 
Size      hjj 

Size 

t* 

Size 

-S 

°O 

Size 

4-» 

in 

.  ^ 

in 

•  J*^ 

in          o 

in 

•  ^ 

in 

in 

3 

Deci- 

jS £ 

Deci- 

Z  ^ 

Deci-     £& 

Deci- 

55.S 

Deci- 

£ " 

Deci- 

mals. 

1 

mals. 

£ 

mals.          £ 

mals. 

mals. 

£ 

mals. 

z 

.413 

H 

.266 

11 

.188 

29 

.134 

47 

.077 

65 

.033 

Y 

.404 

G 

.261 

12 

.185 

30 

.127 

48 

.075 

66 

.032 

X 

.397 

F 

.257 

13 

.182 

31 

.120 

49 

.072 

67 

.031 

w 

.386 

E 

.250 

14 

.180 

32 

.115 

50 

.069 

68 

.030 

V 

.377 

D 

.246 

15 

.178 

33 

.112 

51 

.066 

69 

.029 

u 

.368 

C 

.242 

16 

.175 

34 

.110 

52 

.063 

70 

.027 

T 

.358 

B 

.238 

17 

.172 

35 

.108 

53 

.058 

71 

.026 

S 

.348 

A 

.234 

18 

.168 

36 

.106 

54 

.055 

72 

.024 

R 

.339 

1 

.227 

19 

.164 

37 

.103 

55 

.050 

73 

.023 

Q 

.332 

2 

.219 

20 

.161 

38 

.101 

56 

.045 

74 

.022 

P 

.323 

3 

.212 

21 

.157 

39 

.099 

57 

.042 

75 

.020 

O 

.316 

4 

.207 

22 

.155 

40 

.097 

58 

.041 

76 

.018 

N 

.302 

5 

.204 

23 

.153 

41 

.095 

59 

.040 

77 

.016 

M 

.295 

6 

.201 

24 

.151 

42 

.092 

60 

.039 

78 

.015 

L 

.290 

7 

.199 

25 

.148 

43 

.088 

61 

.038 

79 

.014 

K 

.281 

8 

.197 

26 

.146 

44 

.085 

62 

.037 

80 

.013 

J 

.277 

9 

.194 

27 

.143 

45 

.081 

63 

.036 

I 

.272 

10 

.191 

28 

.139 

46 

.079 

64 

.035 

In  using  the  gages  known  as  Stubs'  Gages,  there  should  be  constantly  borne  in 
mind  the  difference  between  the  Stubs  Iron  Wire  Gage  and  the  Stubs  Steel  Wire 
Gage.  The  Stubs  Iron  Wire  Gage  is  the  one  commonly  known  as  the  English 
Standard  Wire,  or  Birmingham  Gage,  and  designates  the  Stubs  soft  wire  sizes.  The 
Stubs  Steel  Wire  Gage  is  the  one  that  is  used  in  measuring  drawn  steel  wire  or  drill 
rods  of  Stubs'  make,  and  is  also  used  by  many  makers  of  American  drill  rods. 

<I48) 


Geometry 


Geometry  is  the  science  which  teaches  the  properties  of 
lines,  angles,  surfaces  and  solids. 

A  point  indicates  only  position  and  has  neither  length, 
breadth  or  thickness.  A  point  has  no  magnitude. 

A  line  has  length,  but  no  breadth  or  thickness  ;  it  is  either 
straight,  curved  or  mixed. 

A  straight  line  is  the  shortest  distance  between  two  points. 

A  curved  line  is  continuously  changing  its  position. 

A  mixed  line  is  composed  of  straight  and  curved  lines. 

A  surface  has  length  and  breadth,  but  no  thickness  ;  it  may 
be  either  plane  or  curved. 

A  solid  has  length,  breadth,  and  thickness  or  depth. 

An  angle  is  the  inclination  of  two  lines  which  intersect  or 
meet  each  other.  The  point  of  intersection  is  called  the  vertex 
of  the  angle.  An  angle  is  either  right,  acute  or  obtuse. 

A  right  angle  contains  90  degrees.  An  acute  angle  contains 
less  than  90  degrees.  An  obtuse  angle  contains  more  than  90 
degrees. 


FIG.  i.  \  \  FIG.  3 


Right  Angle.  Acute  Angle.  Obtuse  Angle. 

*  Polygons. 

Polygons  are  plane  figures  bounded  on  all  sides  by  straight 
lines,  and  are  either  regular  or  irregular,  according  to  whether 
their  sides  and  angles  are  equal  or  unequal.  The  points  at 
which  the  sides  meet  are  called  vertices  of  the  polygon.  The 
distance  around  any  polygon  is  called  the  perimeter. 

A  figure  bounded  by  three  straight  lines,  forming  three 
angles,  is  called  a  triangle. 

The  sum  of  the  three  angles  in  any  triangle,  independent  of 
its  size  or  shape,  makes  180  degrees. 

All  triangles  consist  of  six  parts ;  namely,  three  sides  and 
three  angles.  If  three  of  these  parts  are  known,  one  at  least 
being  a  side,  the  other  parts  may  be  calculated. 

A  triangle  is  called  equilateral  when  all  its  three  sides  have 
equal  length.  Then  all  the  three  angles  are  equal,  namely,  60 
degrees,  because  CO  X  3  =  180.  ( See  Fig.  4). 

*  Some  authorities  define  as  polygons  only  figures  having  more  than  four  sides. 


GEOMETRY. 


A  triangle  is  called  a  right-angled  triangle  when  one  angle 
is  00  degrees ;  the  other  two  angles  will  then  together  consist  of 
90  degrees,  because  90  -f  90  =  180.  ( See  Fig.  5). 

An  acute-angled  triangle  has  all  its  angles  acute.      (See 


FIG.  4. 


FIG.  6. 


Equilateral  Triangle.  Right-Angled  Triangle. 


Acute  Triangle. 


The  longest  side  in  a  right-angled  triangle  is  called  the 
hypothenuse  and  the  other  two  sides  are  called  the  base  and 
perpendicular.  The  square  of  the  length  of  the  hypothenuse  is 
equal  to  the  sum  of  the  squares  of  the  lengths  of  the  other  two 
sides.  (  See  Fig.  5). 


= 


From  this  law  the  third  side  of  a  right-angled  triangle  can 
always  be  found,  when  the  length  of  the  other  two  sides  is 
known.  Thus  :  (  See  Fig.  5). 


If,  instead  of  the  letters  a,  b,  and  c,  numbers  are  used,  for 
instance,    a  =  3  and  b  =  4  ;  what  then  is  the  length  of  c  ? 


A  square  is  a  plane  figure  having  four  right  angles  and 
bounded  by  four  straight  lines  of  equal  length.  (See  Fig.  7). 


FIG.  7. 


GEOMETRY. 


A  parallelogram  is  a  plane  figure  whose  opposite  sides  are 
parallel  and  of  equal  length.     (  See  Fig.  8). 

A  rectangle  is  a  parallelogram  having  all  its  angles  right 
angles.    (  See  Fig.  <J). 

A  trapezoid  is  a  plane  figure  bounded  by  four  straight 
lines,  of  which  only  two  are  parallel.     (See  Fig.  10). 


FIG.  a. 


FIG.  9. 


FIG.  10. 


Parallelogram. 


Rectangle. 


Trapezoid. 


FIG.  11. 


A  trapezium  is  a  plane  figure  bounded  by 
four  sides,  all  of  which  have  unequal  length. 
(See  Fig.  11). 

Polygons  having  four  sides,  and  conse- 
quently four  angles,  are  usually  called  quad- 
rangles. Polygons  having  more  than  four 
sides  are  named  from  the  number  of  their 
sides. 


Trapezium. 

Thus: 

A  polygon  having  five 

"        "  "  six 

"        "  "  seven 

"        "  "  eight 

"        "  "  nine 

"        "  "  ten 

"  "  eleven 

"        "  "  twelve 


sides  is  called  a  pentagon. 

"  "  "  a  hexagon. 

"  "  "  a  heptagon. 

"  "  "  an  octagon. 

"  "  "  a  nonagon. 

"  "  "  a  decagon. 

"  "  "  an  undecagon. 

"  "  "  a  dodecagon. 


The  sum  of  the  degrees  of  all  the  angles  of  any  polygon 
can  always  be  found  by  subtracting  2  from  the  number  of  sides 
and  multiplying  the  remainder  by  180. 

For  instance : 

The  sum  of  degrees  in  any  quadrangle  is  always  (4  —  2) 
X  180  =  360  degrees. 

The  sum  of  degrees  in  any  pentagon  will  always  be  (5  —  2) 
X  180  =  540  degrees. 

This  is  a  useful  fact  to  remember  in  making  drawings,  as  it 
may  be  used  for  verifying  angles  of  polygons. 


'5* 


GEOMETRY. 


FIG. 


Circles. 

The  Circle  is  a  plane  figure  bound- 
ed by  a  curved  line  called  the  circum- 
ference or  perphery,  which  is  at  all 
points  the  same  distance  from  a  fixed 
point  in  the  plane,  and  this  point  is 
called  the  center  of  the  circle.  ( See 
point  c,  Fig.  12). 

A    Diameter    is    a    straight    line 
passing  through  the  center  of  a  circle  or 
a  sphere,  terminating  at  the  circumfer- 
Circle.  ence  or  surface.     (See  line  e-d,  Fig.  12). 

A  Radius  is  a  straight  line  from  the  center  to  the  circum- 
ference of  circle  or  sphere.  (  See  line  c-f,  Fig.  12). 

Diameter  =  2  X  radius.  The  ratio  of  the  circumference  to 
the  diameter  of  a  circle  is  usually  denoted  by  the  Greek  letter  TT 
and  is  expressed  approximately  by  the  number  3.1416  or 

Thus,  if  the  circumference  is  required,  multiply  the  diameter 
by  3.1416.  If  the  diameter  is  required,  divide  the  circumference 
by  3.1416. 

A  Chord  is  a  straight  line  terminating  at  the  circumference 
of  a  circle  but  not  passing  through  the  center  of  the  circle.  (See 
line0-£,  Fig.  12).  The  curved  line  a-b,  or  any  other  part  of  the 
circumference  of  a  circle,  is  called  an  arc. 

Any  surface  bounded  by  the  chord  and  an  arc,  like  the 
shaded  surface  a-b,  is  called  a  segment. 

Any  surface  bounded  by  a  chord  and  its  two  radii,  like  the 
shaded  surface  c-f-d,  is  called  a  sector. 

PROPERTIES  OF   THE   CIRCLE. 

Circumference  =  Diameter  X  3.1416 
Area  =  (  Diameter)2  X  0.7854 

Diameter          =  Circumference  X  0.31831 
Diameter          =-J^ 


Diameter  =  1.1283  X 

Circumference  =  3.5449  X 


area 


Length  of  any  arc  =  Number  of  degrees  X  0.017453  X  radius. 
Length  of  arc  of  1  Degree  when  radius  is  1  is  0.017453. 
Length  of  an  arc  of  1  Minute  when  radius  is  1  is  0.000290888. 
Length  of  an  arc  of  1  Second  when  radius  is  1  is  0.000004848. 
When  the  length  of  the  arc  is  equal  to  the  radius  the  angle 
is  57°  17'  45"  =  57.2957795  degrees. 


TRIGONOMETRY. 


153 


TRIGONOMETRY. 

Trigonometry  is  that  branch  of  geometry  which  treats  of 
the  solution  of  triangles  by  means  of  the  trigonometrical 
functions. 

When  the  circumference  of  a  circle  is  divided  into  360  equal 
parts  each  part  is  called  one  degree. 

One  fourth  of  a  circle  is  90  degrees  =  right  angle,  because 
4  X  90  =  360.  ( See  Fig.  13.) 

FIG.  13.  fc 


1  degree  =  60  minutes  (60'.) 
1  minute  =  60  seconds  (60".) 


cotangent  (cot.) 


FIG   14. 

Circle  =  4  right  angles. 
Circle  =  360  degrees  (360°.) 

Concerning  the  angle  n  (see  Fig.  14)  the  following  are  the 
trigonometrical  functions : 

c  g  radius  =  1.  c  h  cosecant  (cosec.) 

d  b  sine  ( sin.)  g  f  tangent  ( tan.) 

c  d  cosine  (cos.)  k  h 

c  f  secant  (sec.) 

The  complement  of  an  angle  is  what  remains  after  sub- 
tracting the  angle  from  90°.  Thus,  the  complement  of  an  angle 
of  30°  is  60°  because  90  —  30  —  60. 

The  supplement  of  an  angle  is  what  remains  after  subtract- 
ing the  angle  from  180°.  Thus,  the  supplement  of  an  angle  of 
30°  is  150°  because  180  —  30  =  150. 

As  all  circles,  regardless  of  their  size,  are  divided  into  360 
degrees,  the  trigonomical  functions  must  always  be  alike  if  the 
radius  and  the  angle  that  they  denote  are  alike. 

It  is  on  this  basis  that  the  tables  of  trigonometrical  func- 
tions are  calculated,  and  as  radius  is  used  the  figure  1 . 

In  Table  No.  20,  the  natural  sine  of  30°  is  given  as  0.5;  this 
means  that  if  the  line  c  g  (see  Fig.  14)  is  1  foot,  meter,  or  any 
other  unit,  and  the  angle  n  is  30  degrees,  the  line  db  will  be  0.5 
of  the  same  unit  as  the  line  eg. 

Sine  45°  =  0.70711  ;  that  is,  if  the  the  angle  n  is  45  degrees 
and  the  line  c  g  is  1  of  any  unit,  the  line  d  b  is  0.70711  of  the 
same  unit. 

Cos.  30°  ==  0.86603 ;  that  is,  if  the  angle  n  is  30  degrees  and 
the  line  c  g  is  1  of  any  unit,  the  line  a  b  or  c  rf  is  0,86603  of  the 
same  unit, 


TRIGONOMETRY. 


Sec.  30°  =  1.1547 ;  that  is,  if  the  angle  n  is  30  degrees  and 
the  line  cgis  1  of  any  unit,  the  line  cf  is  1.1547  of  the  same  unit. 

Cosec.  30°  =  2;  that  is,  if  the  angle  ;/  is  30  degrees  and  the 
line  eg  is  1  of  any  unit,  the  line  c h  is  2  of  the  same  unit. 

Tang.  30°  =  0.57735 ;  that  is,  if  the  angle  n  is  30  degrees 
and  the  line  c  g  is  1  of  any  unit,  the  line  g  f  is  0.57735  of  the 
same  unit. 

Cot.  30°  =  1.73205 ;  that  is,  if  the  angle  n  is  30  degrees  and 
the  line  c  g  is  1  of  any  unit,  the  line  k  h  is  1.73205  of  the 
same  unit. 

Increasing  the  angle  n  will  increase  sine,  tangent  and 
secant,  but  will  decrease  cosine,  cotangent  and  cosecant. 

When  the  angle  n  approaches  90°,  the  tangents  g  f  increase 
more  and  more  to  infinite  length.  When  n  actually  reaches  90° 
of  course  c  b  coincides  with  c  k  and  becomes  parallel  to  gf,  so 
that  in  an  angle  of  90°  both  /the  secant  and  the  tangent  have 
infinite  length,  which  is  denoted  by  the  sign  oo,  and  cosine  and 
cotangent  have  vanished. 

In  the  first  quadrant  (that  is  when  angle  n  does  not  exceed 
90°)  the  trigonometrical  functions  are  all  considered  to  be 
positive  and  are  denoted  by  +  (plus).  When  the  angle  n 
exceeds  90°,  only  sine  and  cosecants  remain  positive ;  all  the 
other  functions  have  become  negative  and  are  denoted  by  — 
(minus). 

The  following  table  gives  the  properties  of  the  trigono- 
metrical functions  in  the  four  different  quadrants : 


Degree. 

Sine. 

Cosine. 

0°  to    90° 
90  °  to  180  ° 
180°  to  270° 
270°  to  360° 

Increase   from  0  to  radius  -j- 
Decrease  from  radius  to  0  -j- 
Increase   from  0  to  radius  — 
Decrease  from  radius  to  0  — 

Decrease  from  radius  to  0  -|- 
Increase  from  0  to  radius  — 
Decrease  from  radius  to  0  — 
Increase   from  0  to  radius  -f~ 

Degree. 

Secant. 

Cosecant. 

0°  to    90° 
90°  to  180° 
180°  to  270° 
270°  to  360° 

Increase  from  radius  to  00  -f- 
Decrease  from    OQ  to  radius  — 
Increase   from  radius  to  °°  — 
Decrease  from    Oo  to  radius  + 

Decrease  from  O°  to  radius  -f- 
Increase   from  radius  to  °°  -j- 
Decrease  from   0°  to  radius  — 
Increase   from  radius  to  0°  — 

Degree. 

Tangent. 

Cotangent. 

0°  to    90° 
90°  to  180° 
180°  to  270° 
270°  to  360° 

Increase   from  0  to  O°  -j- 
Decrease  from  O°  to  0  — 
Increase   from  0  to  °°  -f- 
Decrease  from   O°  to  0  — 

Decrease  from  Oo  to  0  -f 
Increase   from  0  to  00  — 
Decrease  from  O°  to  0  -f- 
Increase    from  0  to  °0  — 

From  the  rule  that  the  square  of  the  hypothenuse  is  equal 
to  the  sum  of  the  squares  of  the  base  and  the  perpendicular,  it 
also  follows  that: 


TRIGONOMETRY. 


155 


Sin.2  +  cos.2  =  radius2. 
Tang.2  -f-  radius2  =  secant2. 
Cot.2  +  radius2  =  cosecant2. 

But  the  trigonometrical  tables  are  calculated  with  radius 
1,  hence, 

sin.2       -fcos.2=l. 
tang.2     +1       —  sec.2 
cotang.2  -j-  1       —  cosecant2. 

tang.     = 


secant  = 


cotang.  = 


cosec.  = 


cosm. 
1 


cosin. 
sin. 

1 


cotang.  — 


cosin.    = 


tang. 

-   Vl  —  sin.5 


tang.     = 


secant  = 


sin.       = 


sin. 


sin.        = 


cosm.    = 


FIG.  is. 


FIG.  16. 


FIG.  17. 


Sine  and  Cosine  of  the  Sum  of  Two  Angles. 

(See  Fig.  15). 

Sine  (a  -f  b}  =  sin.  a  X  cos.  b  -f-  cos.  a  X  sin.  b. 
Cos.  (a  +  b}  =  cos.  a  X  cos.  b  —  sin.  a  X  sin.  b. 

Sine  and  Cosine  of  Twice  any  Angle. 

(See  Fig.  16). 

Sin.  2  a  =  2  X  sin.  a  X  cos.  a. 
Cos.  2  a  =  cos.2  a  —  sin.2  a. 

Sine  and  Cosine  of  the  Difference  of  Two  Angles. 

(See  Fig.  17). 

Sin.      (a  —  b}  =  sin.  a  X  cos.  b  —  cos.  a  X  sin.  b. 
Cosin.  (a  —  b]  =  cos,  a  X  cos.  b  -f-  sin,  a  X  sin,  b, 


s 


8 


"> 


. 


t 


s 


ft 


8 


t 


TRIGONOMETRY.  157 

The  Trigonometrical  Table  and  Its  Use. 

Table  No.  21  gives  sine,  cosine,  tangent,  and  cotang,  to 
angles  from  0  to  90  degrees  with  intervals  of  10  minutes. 

For  sine  or  tangent  find  the  degree  in  the  left-hand  column 
and  find  the  minutes  on  the  top  of  the  table.  For  instance, 
sine  to  18°  40'  =  0.32006. 

If  cosine  or  cotangent  is  wanted,  find  the  degree  in  the 
column  at  the  extreme  right  and  the  minutes  at  the  bottom  of 
the  table.  For  instance,  cotang  48°  10'  =  0.89515. 

As  the  table  only  gives  the  angles  and  their  trigonometrical 
functions  with  10-minute  intervals,  any  intermediate  angle  must 
be  calculated  by  interpolations.  For  instance,  find  sine  of  60° 
15'  10". 

Solution : 

Sine  60°  20'  0"  =  0.86892 

Sine  60°  10'  0"  =  0.86748 

Difference  of    0°  10'  0"  =  0.00144 

60°  15'  10"  —  60°  10'  0"  =  0°  5'  10'  =  310  seconds  and  a 
difference  of  10'  =  600"  increases  this  sine  0.00144.  Therefore 
a  difference  of  310  seconds  will  increase  the  sine. 

310  *t£»M  =  0.00074 
and  sine  60°  10'  0"  —  0.86748 

Therefore  sine  60°  15'  5"  =  0.86822 

IMPORTANT. — During  all  interpolations  concerning  the 
trigonometrical  functions,  remember  the  fact  that  if  the  angle 
is  increasing  both  sine  and  tangent  are  also  increasing,  and 
corrections  found  by  interpolations  must  be  added  to  the  num- 
ber already  found ;  but  as  the  cosine  and  cotangent  decrease 
when  the  angle  is  increased,  for  these  functions  the  corrections 
must  be  subtracted. 

Interpolations  of  this  kind  are  not  strictly  correct,  as  neither 
the  trigonometrical  functions  nor  their  logarithms  differ  in  pro- 
portion to  the  angle.  The  error  within  such  small  limits  as 
10  minutes  is  very  slight.  When  very  close  calculations  of 
great  distances  are  required,  tables  are  used  which  give  the 
functions  with  less  difference  than  10  minutes;  but  for  mechan- 
ical purposes  in  general  these  interpolations  are  correct  for  all 
ordinary  requirements.  It  is  very  seldom  in  a  draughting  office  or 
a  machine  shop  that  any  angle  is  measured  for  a  difference  of 
less  than  10'. 

To  Find  Secant  and  Cosecant  of  Any  Angle. 

Divide  1  by  cosine  of  the  angle  and  the  quotient  is  secant 
of  the  same  angle. 

Divide  1  by  sine  of  the  angle  and  the  quotient  is  cosecant 
of  the  same  angle. 


I  to 


TRIGONOMETRICAL    TABLES. 


00  00  00  CO  00  00  CO  00  CO  OO  £—  L—  t—  t— 


s 


—  . 

r-io?iOOOOO'Mr7Ot^O: 


g 


>Q  G:   -f  X   ?M  CO   c:    -._i  O  I-  Cl  O  —  •— >  O  00  »Q  <M  00  C4  CO  Ci  O 

^H07^*;5xO'-^-::t.'..'t^ooOCMr7>Or-ccO'^o7)Oi^co 
opoo<^'— i  —  i— i  — .I— i  —  <Ncxq7>iiMCNicN|cccoeococcco 
bbobbbo'bbbo'bbbbbbbbbbbb 


CM  -^  »O  t-  OS  »—  i  C<l  T*<  CO  t^  OS  i—  (CCi^COOOOS'-HCO^COt- 


ecOGOfMCOO^t— 
Ol^iOOlO 
C<lCOiOt—  Oi 


COiQCOt—lr-.t—CO'^T—  (t--CQt—  OCO 
OS^OCQOt—'^'—  tCOiOi—  iGO^i—  It- 
iOlr—Oi'—  iC^-^COl.--Oii—  ICM-ffCDtr- 


<Mt—  (Mt-r-l 


i—  (COOCOCOO(T>lCOvOlTC3iO(M"*iOj>- 


TRIGONOMETRICAL    TABLES. 


'59 


CD»O^COG^i-»OO500r-CDO^CO(Mi— lOOiOOt—  «DiO 


CO*O^CO<Mi-»OC500r-CO*Q 
CO  CO  CO  CO  CO  CO  CO  iQ  *O  *Q  *Q  ^O 


^*    ^1    h-»    O5   !>•   ^H   <^    i¥ 

t-COCOCr.  ^OOOOOSCD^Of-OOCDCO^O^OCO^ 

coc^oocoos^ooos^aseor-i— iiooscMcoosoj^t^ 
coocoooo^cvi^Kr 


*O  CO  CD  CO  CO  CO  CO  CO  t>» 


^  CC  i~^  CO  CO 


tOCOCOCDCO 


I 
I 


O  00  o:  t-  O  OS 
O  OO  O  T-I  CO  OO 


CO-^^iOOOi— lOOiC^liO 


CO  ^  CV|  t^  OS  t-  i— i 

t-  t-  CD  CO  Oi  •*  00 
O  CO  <M  00  CO  OS 
O5  O  CM  CO  O  CD  00 


I-H  iO  t"»  00  CO  CO  Ir^- 


i6o 


TRIGONOMETRICAL    TABLES. 


•»*  CO  <M  i—  l 

•^•^•^^ 


cocoi— ii>-o  —  ocob— 'Ocoo— 'bcociot— cooiO"— i 

?  ^  ^  t^  ^  t^-b*  r~  CO  00  00  0(3  00  <$  OO  00  00  00  00  0»  C9  0»  Oi 


<Mt—  OOt-T—  i 


S 


SO  O  OC  ^^  00  O  t~  t>»  *""*  GO  00  C^l  GO  ^O  CO  OS 

GO  C^  O  *~^  ^J  ^C  "*^  »O  CO  t-—  OO   ^C  O  O  —  '—"'  C^l 

t^^i>-i>:t-oqcooqoqGOGOoooqoqcooocnoiOiOi 
'boo' 


COiOt-GOOi— ICflCOCO-^^COCOC^^HOOOt^-TTC^Oir^Ot 

XGOOOOOGOXCOGOOOOOCiOiO^OJ 

bob 


1-1  t—  r-i  CO  »-  t-OiOO(M(MOOOOCO(MiO(MCOCDCOeCOOCO 


8 


Me9TiQ«o££o0S 

bob 


TABLES. 


00  A  C4  <O  i-*  CD  O3  O.t-**  tQ  CO  i-4  Od  t—  «O  93  O  v  CO  OB 


O-t-  »o  eo  i-»  o»  t-  »o  W  o  co  co  o»  o 

rr  co  i— i  -r  x  co  -M  >o  o  -M  >c  o  cc  co 


—  SO  <— i  iQ  O  -*<  00  i~ i-rl^ 


Ci  C^>  —  -  —  -  *  —  *  C^  —  -  CT-  *-  —  -  O^  —  -  — - 


Ct  Ct.  Ct  O  C^  —  -  OS  —  -  -T-  -- 


§<NiOGO<^-*t«t^O<M-'f<£>GOai 
Oi—'— I«OI-H?DC:  a-^r^t-  ^ 


01  os  oi  as  os  os  os  o:  o»  a ;  os 


TRIGONOMETRICAL   TABLES. 


00  00  GO 


CO  (M  i— lCOCiOas-"fCCCOGOCOl^COOiQCO<MCOt~COCOGO 
•^Ol'^'Oi-*''— 'l>->OCOCCCO>OOOCOOit-l^-OiCOO5GOO^ 
lr-^(MO5t-O<MOOOCO"*C^OO5t^CCiO^'*|COOt-f'* 
>-iCOiOCOGCO<N^iOt^aii— iCO-'tiCOGCOC<J^f|;OGCO<M 


I 

I 


00  CO  CO  i— lOOcCCO  — 
OOOOOO^H^^22§S 


2 


TRIGONOMETRICAL   TABLES. 


'63 


32SS 


i-iGOCSOTt<CSGOC^C*lGO 


r- lOCDOSl— iT-Hl— IOOW 


GO-*fOC<li>-COC<l.t>-'— 'Jt- 
<M  (M  »0  IM  <M  t-  CO  O:  GO  ^H 


W 
en 

a 


^r -CSt-CO 


o  o  o  o  o 


o  o      o  o 


COCO— 'COCO*- l^ 


I 


CO'*>0«Or-G005O'-HiMCO^»0«01r-GOOSO— «<MCO^ 

<M!M(MC^!^!MCMCCCOCOCOCOCOCOCOC050"*'^'«11'*'^ 


164 


1'ABLES. 


2 


O"fiOCOOOr--*00 

OCCiOCOr-OSC:OCO—  •  »Q  00  CO  CM  O  O  1—  CO  CO  iO  O  00  O 
CCiOCi^JClO^(M-r   i  (M  O  •«*!  «  O  O 


OCMOO  —  "* 


OT-  i  O  i—  1  -fi 

—  CO  CO  iO  O  0 


Ol^"*CO^ 

p«  «  o  •^'-oiJ 

OOr-ir—  T—  t 


t-T— i  zO  T— it-COOtiQC^CS^^COC^TMCOiO 
r^T^rHrtf^rHT-ir-Jr-iT-ir-i'T-ioic<icsioqC^ 


lO^O^t-iMi—iT-l 
OCT.  t-Cit^COt^T—  I 
OOOit-<D 
CMiOOSCO 
OOOi—  i 


OOOr—  IT—  i 


CO  Ci  O  p  i—i  M  ^ 

i-i  i-J  c>i  c^i  c^'  <r>i  c<i 


O  T- i  (M^  CO 

w'  ci  IM'  w 


O^t~T— iiOOS'^COCOGOCOCJT'— i  t-  T^  »— i  OS  t^-  ?C  COCOr- 
OOO^— I  T— (  r- iC^C^COCO'^TTiOCDsOt-GCOOOO'— 'OQCO 


§ 


-i  <rq  <M'  <H  (H 


COlr-i—  it—  tOO 
OCCOCOt—  Ol 
iO(MOO  —  '* 


tRKJONOMETRiCAL   TABLES. 


^H  _l  ^  CM  0 


i-i  *-  •005'*—  <CliO^-i»OM<OOCOeO<MOCO"*OOOOi-l 
CD  CO  O  •«  rl  *O  O  CO  t~  S4  00  CO  O  C9  O  CO  -^*  1>>  O  49  00  t« 


T—  i  O  i~"i  "^  '""'  '—  ^  t->-  GO  00 


§ 


s 


OCVICDC^OO^ 
i—i  r-  1  i—  4  C<I  CO  •—  « 


I-H  ri  1-1  <M  CO 


1^<M'—  i  CO 


i-l>*O-'CO 
—  rH  CM  CO  CO 


GO 
-<* 
t~ 
-^ot^t^coco 


ooccoos 

CCOCO<M 
'^OSCCI^ 
r-l  rt  W  iO 


166  TRIGONOMETRY. 

Logarithms  Corresponding  to  the  Trigonometrical 
Functions. 

Table  No.  22  gives  the  logarithms  corresponding  to  sine, 
cosine,  tangent  ana  cotang.  for  angles  from  0  to  00  degrees, 
with  intervals  of  10  minutes.  For  sine  and  tangent  find  the 
degree  in  the  column  to  the  left  and  the  minutes  at  the  top  of 
the  table.  For  instance : 

Log.  sine  19°  30'  =  9.523495  —  10. 

This,  of  course,  is  also  logarithm  to  the  fraction  0.33381, 
which  is  sine  of  19°  30'. 

For  cosine  and  cotang.  find  the  degree  in  the  column  to  the 
extreme  right  in  the  table,  and  find  the  minutes  at  the  bottom 
of  table.  For  instance : 

Log.  cotang.  37°  10'  =  10.120259  —  10  =  0.120259. 

NOTE. — In  this  table  the  index  of  the  logarithm  is  increased 
by  10,  therefore  —  10  must  always  be  annexed  in  the  logarithm. 

Logarithms  to  angles  between  those  in  the  table  may  be 
obtained  by  interpolations.  For  instance,  find  log.  sine  25°  45'. 

Solution : 

Log.  sine  25°  50'  =  9.639242  —  10 

Log.  sine  25°  40'  =  9.636623  —  10 

Difference  0.002619 

This  difference  in  the  logarithm  corresponds  to  a  difference 
in  this  angle  of  10  minutes  ;  therefore  a  difference  of  5  minutes 
in  the  angle  will  make  a  difference  of  0.001309  in  the  logarithm. 
Thus : 

Log.  sine  25°  40'  =  9.636623  —  10 
Difference       5'  =  0.001309 
Log.  sine  25°  45'  =  9~637932  —  10 

EXAMPLE  2. 

Find  angle  corresponding  to  logarithmic  sine  9.894246  — 10. 

Solution : 

In  the  table  of  logarithms  of  sine : 

9.894546  —  10  corresponds  to  51°  40' 
9.S93544  —  10  corresponds  to  51°  30' 

Difference  0.001002       ~~  corresponds  to   0°  10' 

To  logarithm  9.894246  —  10  must,  therefore,  correspond  an 
angle  somewhere  between  51°  30'  and  51°  40',  which  is  found 
thus : 

The  given  logarithm  is  9.894246  —  10 

Nearest  less  logarithm  9.893544  —  10  for  51°  30' 

Difference  0.000702 

Therefore,  the  correction  to  be  added  to  the  angle  already 
found  will  be : 

0.000702  X   10  _      0 
0.001002 

Thus,  the  logarithmic  sine  9.894246  —  10  gives  51°  37' 


TRIGONOMETRV.  167 

EXAMPLE  3. 

Find  log.  to  tangent  of  50°  45' 
Solution : 

Log.  tangent  50°  50'  =  0.089049 
Log.  tangent  50°  40'  =  0.086471 

Difference       0°  10'  =  0.002578  in  the  logarithm.     There- 
fore a  difference  of  5'  in  the  angle  will  give  0.001289  in  the 
logarithm. 
Thus: 

Log.  tangent  50°  40'  =  0.086471 

Difference  5'  =  Q.QQ1289 

Log.  tangent  50°  45'  =  0.087760 
EXAMPLE  4. 

Find  the  angle  corresponding  to  log.  tangent  9.899049  —  10. 
Solution : 

Log.  tangent  38°  30'  =  9.900605  —  10 

Log.  tangent  38°  20'  9.898010  —  10 

Difference       0°  107  corresponds  to  0.002595 
The  given  logarithm      =  9.899049  —  10 
Nearest  less  logarithm  =  9.898010  —  10  gives  38°  20' 
D  iff  erence  =  0.001039 

The  difference  to  be  added  to  the  angle  already  found  will 

0.001039  X^lO    _ 
De  0.002595  4  ' 

The  tabulated  logarithm  9.898010  —  10  gives  angle  38°  20' 
Difference  0.001039  gives  angle          4' 

Logarithm  9.899049  —  10  gives  angle  38°  24' 

To  Find  Logarithm  for  Secants  and  Cosecants. 

Logarithm  for  secants  is  found  by  subtracting  log.  cosine 
from  log.  1. 

For  instance,  find  logarithmic  secant  30°. 
Solution : 

Log.  1  =  10.000000  —  10 
Log.  cosine  30°  =  9.937531  —  10 
Log.  secant  80°  =  0.062469 

Logarithm  for  cosecants  is  found  by  subtracting  log.  sine 
from  log.  1.     For  instance,  find  logarithmic  cosecant  35°. 
Solution : 

Log.  1    =  10.000000  —  10 
Log.  sine  35°  =    9.758591  —  10 
Log.  co-secant  35°  =    0.241409 

NOTE. — What  is  said  concerning  interpolations  of  trigono- 
metrical functions  in  general  in  the  note  headed  "Important" 
on  page  157,  will  also  apply  to  their  logarithms. 


t6S 


TABLES. 


OSOOt-COOrJ-COfMTHO 
00  00  CO  CO  CO  CO  CO  CO  GO  CO 


OOGOOOiOC^CaOOiOTOCOiOOOOCOOSCodsOSOOCOiOCO 
Tt<  TH  i— I  -^  -rt<  i— (OO^OSCOOOi-HiOOO  —  ^COOprHCOiCt-Cft 
GO  GO  GO  GO  GO  OS  OS 

OOodcOOOGCOSOSGSCSOSOSOSCSOSOSOSOSOSOSOiOSOSOS 
Q 

B 
c/2 

,2n  00  OS  CO  CD  CO  *O  00  CO  £•*••  CO  CO  CO  CO  i"H  CO  QC  CO  *-H  *rJH  «rf  CO  CO  CO 

^^    ^^  t^»  O^  ^O  Tt<  T-H  CO  ^O  Oi  1^-  O  C^  *O  GO  00  CO  CO  CO  T^H  OO  ^  "^*  C*^ 

"osososososososososoi 

kJ  ^t-^r-IOOSiOCM'^IC^'-HOSOS 

iOt^CO»— lOO^CMOSCOOSCOOSOSOOOOr-liO'>HCO-HCOlOJ>" 
^-»  J>»  t^*  ^O  CM  C^l  CO  OS  i~H  O»  t^*  CO  *Q  00  CO  CO  O  *"^  CO  OS  OS  00  !>• 

<M 

^ooco'GOcdodosos'os'osososososososososososososos 

l~'  _         

Jt**  GO  GO  CO  00  GO  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  O^  OS  OS  OS  OS  OS  OS 

OOSOOCOiO^iOfMOOSOSOOOCOOOOC^fMC^OSiO 
ScOOOOOOC^iMOOiOCOCOiOOOOCCOSCOOSOSCOOCOO 

®  |>*^Hi— Irfi^tii-HOO^OSCCCX;  —  >OOO'— i^COOO-- iCO*Ot— 

I    C<t  »O  t--  00  OS  O  O  r-J  ^  CJ  CM  CO  CO  CO  •<*  -*  TT  -^  iO  O  >O  O 

GOGOcdcOCOOSOSOSOSOSOSOSOSOSOSOSOSOSOSOSOSOS 

£  T-lr-I^HrHr-lT-(^-li-iT-(T-((MC^lM 


TRIGONOMETRICAL    TABLES. 


CC  GO  C^  t~-  O".  »— I  O  O5  O  OS  <M  —  OS  CO  C<1  S^  b-  CO  —  O?^H»O 
— (^"tiTOr^t^CO'— i  O  CD  O:  — i  CD  -<f  r^  CD  -r  -— iGOI-00 

CC  OS  00  O  CD  iQ  OJ  GO  Tl   —   iC   O  <M  -^  CC  X  O  OS  -C  I-  I r 

OStO^-lr-'— i  O  CO  —  T^I-OCOSOSO:GOOOCDiCCC"OS 
OtMtiOt-CCO:'— iC^rt-r>OcDt--OOCr.  o^^loT-T'ti 
-  CD  CD  CD  CD  CD  tD  CD  !>•  t"^  r~-  t-*  Ir—  t—  t^-  t~-  r—  00  00  GO  GO  00  00 
os'  os'  os  os  o;  oi  OS  os'  OS  os  os'  os'  oi  Os'  OS  os  os  cs  os'  os'  05  os' 


g 


O  C^  O*  OS  O 


*-;  GO  GO  CO  00  00  00 

C^  Ci  Cw  C^  Gw  G^ 


"*^  CD  O  00 


=0 


05 

I 

I 


JD 
^a 
J 

! 


OOSOSOSOSOSOSOS 


_.....     ,    OJ—  GOOi— i(M'..     , , 

»OCDCOtDCOCpCDt-lr-;I>;l^l-'t^l>;t^t^OOCOOOoqOOGO 


OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS 


170 


TRIGONOMETRICAL    TABLES. 


COCOCOCOCOCOCOCQCQCO^iMiMIMlMS^iMlM 


*«•  *•»  j»  Q  ^  a*  w  o&  ao  »p  •*  •—  O 

COiMr-00»OOCO"1*'iQCOt--Ci<M«OC'3'—  i  CO  00  CO  l~  7  7M  O 
O  i—  i  O  I—  (M  'O  »O  CC  C:  CO  O  »O  -*  O  «C  X  O:  00  :O  CM  1^  O  — 
SO  ^  «-<  t-  ^  O  !O  C«  I—  *7  00  «  00  *5  t  --  >O  o:  CO  i-  O  -r  t-- 


CCt—  I--  00  OS  C5  O  O  —^  •  — 


i—iCDCOOO—  ' 

rH  »r  «  j»  os'r-  ^  o»  w  t~  ^^  ®  w 

l^i^CS^  —  ^fOCCO-tl^l-^CCl  --  (M^IOO—  ' 

»o  ?»  Ci  «o  eo  <?i  »o  i-i  t-  e*  r-  e*  t-  <?»  so  "-1  »o  os  «  «o  o  e 

' 


os  os  o»  os  o  o>  o»  os  os  os  os  os  os  oa  os  os  a>  o  o> 


1 


OOOOr-(T-l-«ti?DCCi—  c-<* 

OOiOGOt—  i^l-^'TCOT—  IQOO-^CCCCCDO    XOO-HOOCS-^CT 

M<t-l^»O':-i-rOrr''-H>OOOO5CCiOO^iOO^OiOCl'— 

-^-—  (XiiOJMOO'*OCDT-H^)^-.CDi—  I  tO  O  "if  OC  <M  CO  Oi  !M« 

>C)CDOt^OCX)C5OO^'—  (M  r^cCCO^^^OiO'O^OC 

00  00  00  y<  00  OQ  00  Ol  OS  OS  OS  OS  OS  OS  OS  OS  OS  Ok  Oi  OS  O>  OS 


.    —  COOS'— i  C  t^  CO  CC  00 

iQ   CC   CC  l>-  3C-   CC   O^   OS  O  '—I  i— •<  C^  7^  ^*   ^  ^  ""^    ^"'  ^"  >LV 

GC  00  X  X  CC  00  00  JO  OS^  OS,  OS  OS  C:   OS  O:   Oi  OS  OS  OS  OS 

oios'os'osos'osososos'os'os'osososos'ososoios'oioioioi 


f-  Q  O  iQ  09  £*  CO 


i— iC^COOOiMCS^OOOSOSCOiOCOO 
os  os  os  os'  os'  os*  os  os'  OS  o;  os  os'  o»  os  OS  os  os*  os  os'  os  os  os'  oi 


--  i 

OCCiO^MGOiO—  it-COGO-fCr.^O^CCX 
iO  iO  CD  t-  I-  OC  Oi  O5  O  O  —  i  •—  C^J  (M  CO  CC 

ooocxoooocooooooiosoioic;  osaij. 


.  °:  ^ 
OS  OS  Oi 


TRIGONOMETRICAL    TABLES. 


171 


s  s 


OS  00  t>»  CO  *Q  "^  CO  <?5  T-H 


^^  c^  *Q  do  o  c^  ^  co  oo  ^2  *"^  co  ^  *o  co  t^»  06  oo  os  os  os  ^^    ^^ 

!>•  t>-  t^  00  00  GO  00  GO  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  <""">      ^^ 

OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  O»  OS  OS  OS  OS  OS  OS  OS  OS  O* 
OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  O*  OS  OS  OS  OS  OS  OS  O 

CO  ^  ^^  !>•  ^^  OO  CO  CO  OS  C^  r-H  t~*»  OS  QO  CO  >O  TO  DO  ~-* 

*O        cpt^-t^r- ocooooooo6ososososososo;osososososa& 

OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS 

c 

OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  OS  £3 

ai 

d 

CO  eor^i—  |>.i-.  OOOOOOOOOOOSOSOSCT-OSOSOSOSOSOSC:  OS  CO  £ 

•fi 

2 

COCOt^(MOSi-HOOCOCOQ^OOSCOOCO(MOOOOSCO'— I 
T-^  T-^  iju  »i^j  <^^j  i^v  u*;  ^f  ILJ  CT^J  w«>  "^r  r^-  <^^  ^^  ^^  GO  >O  ™^  *O  00  CS 

W     CO  !>•  l>-  t^«  Is*  00  00  QO  QO  00  OS  OS  OS  OS  OS  OS  OS  ui  OS  OS  OS  OS 

OJ  OS  OS  OS  o:  OS  05  OS  OS  OS  OS  os  OS  o;  OS  OS  OS  OS  OS  OS  O:  OS 
OS  OS)  OS  OS  OS  OS  OS  OS  O  OS  OS  OS  OS  OS  OS  O^  OS  OS  OS  OS  OS  OS 

OS  OS  OS  OS  OS  OS  OS  OS  O^  O^  OS  O^  OS  OS  OS  OS  OS  w  -  OS  OS  OS 
Oi  OS  OS*  05  05  OS  OS  OS  OS'  OS*  OS  OS  OS  OS*  OS  OS  O:  OS*  OS  Os'  OS*  OJ 

tr~»  **—s  ^4  iw  uu  <^  u^  'n1  ^^  -*~-  ^  "^  CO  *^  »C  CO  t^*  GO  00  C^  OS  OS 

^•^     CO  !>•  t'-  t*"  !>•  00  OO  QO  00  00  O^  Oi  OS  OS  OS  OS  OS  OS  OS  OS  OV  OS 

OS  OS  OS  OS  OS  OS  O*  OS  OS  OS  OS  OS  OS  CS  OS  OS  OS  OS  OS  OS  OS  OS 

OS  OS  OS  OS  OS  OS  OS*  OS  OS  OS  OS  OS  05  OS  OS  OS  OS  OS  OS  OS*  OS  OS 


172 


TRIGONOMETRICAL   TABLES. 


1 


3 


(M  CO  <M  O  >O  OS 


CO>O 


ft 


T-IOiCDO5Oi»OOiiOOieCi—  li-H 
»OCCiOT—  i>O-*CT.  O—  it-00-* 


i^-  GO  GO  GO  GO  GO  O^  O^  C^  OS  OS  OS  OS  C^  OS  OS  OS  CS  OS  OS  OS  OS  OS 


s 

^ 

•B 

! 


t-OOQOOOCOGOOSOSOSOSOSOSOSOSOSOSOSOSOSOSOSOSOS 


82S 


OSO.OSOSOS 


g^S 


TRIGONOMETRICAL   TABLES. 


173 


COt-COCDI^OCCl^CXDr-<aOCs3CDi--t^CDi-HCDCOiOCO 

—H  -*a*   CD   QO 
01^  C?^0  £> 

CO  lO  CO  OO  Q  ^^ 

i-t-t-r-~t-t-.ao6ooboooooooiosasciO5 

OS  OS  OS  05  01  O 

!>•  CD  CO  CO  <M  CO  ^-i 

i™^  CO  OI  O  CO  t—  O  w»  w»  i—  <«N  ILJ  <i^  "^  t.xi  I-—  ti>i  vv  uu  T-^  v«j  ^r  •% 

OS  O5  OS  Ol"  OS*  OS*  OS  O5  O5*  OS  O5'  OS  OS  OS*  Ol'  O5  OS*  OS*  OS  Ol*  O5  O5 

"  -2 

x^*  *^  i~*~  yu  >u)  i>»  >i«?  <*->  i"^  W3  icj  iju  oa  ou  u^>  ^^  (f  <*->  crj)  113  1^-  O^ 
C*&C*a*&ci&&&C*C*G*0*&G*&0*C*aio:>G*Gi 

O5  CO  CM  O  O 

o  o^  c>  *o  ^o  w        ^2 

^^     -_  -   --   ,-   —    —  .  ..^-^^  T^  ^  CO  O  CQ  ^J*     •* 

OO     CO  iQ  t^  Oi  *"H  CO  O  lr^-  00  ^^  C^l  CC  O  iO  00  ^^  ^™  CO  ^  CO  tr-*  Oi     CO     Kr\ 

CDcDcqcDt-.  t-lr-t-t-00000000000005050105050505 

a 

OOt^COCOCDOiiOOiCOOi^CO^COOOiOC<ICOOCO  O 

^CC^CO^cXOl^OiCOC^OiCDOr-iC^^CD^C^Oi 

_  ^^  ^H 

CD  ^D  CD  l^  t^-  t^»  !>•  1-"  00  00  00  CO  GO  CO  OO  Oi  Oi  Oi  Oi  Oi  Oi 

OS 

h3 

91?  *^  *^T  P?  T~^  ^  "^  ^  ^O  Oi  ^  CO  ^t1  CD  t~^  Oi  T*^  ^-1  ^J*  *O  t^*  OO     O 

b 

OS  os"  o:  os  oi  oi  o:  OS  oi  oi  oi  oi  OS  oi  oi  oi  oi  OS  os'  oi  oi  os 


174 


TRIGONOMETRICAL   TABLES. 


t-«DiO-*eO<M-HOO500t^SD>O^CO(M 


CO^COt-?Oi-<OC005COCOCOr-lCl 

cD^cDcooocoo:oocot^  —  co— !•: 


CD  -*  CC  I—  t-  00  O 

!M   Tt   — I    M   r-   -*   OS 


—  J^  CM  00  ~fi  —  lr-  rt<  — 


O  O  rH   r-t   .—  r-  1   —,   — 


ooooooc 


~  CO  ,O&  »O  rri  00   O  3*  O  00 


OO(M'-H«OvOO'—  i 


^OOO7<I^fCDOO 
C      (M  CO  CO  CO  CO  CO 


COQO'*!—  (O^—  l^O 
i-<;OCV100Tf<OCOCO 


O:cC>-^<MOC10Gt—  OOOS 


ooooooooc> 


<X>—  >t--<MOO-r—  i 


TRIGONOMETRICAL   TABLES. 


J75 


8 


S 


I 


O5^00C^QOt--i—  Ii-Hi—  I  OS 

§COCO>-i 
<M-^t^ 


CO—  ilr^"^Tt<Ol 
(MCOOO^-xOr-ii 


t-t^aO<MOOOO«-"OOr-  IT—  l 


1^*  lr-»  00  GO  GO  CO  GO  GO  GO  GO  GO  GO 


s 


s 


,76 


TRIGONOMETRY. 


Solutions  of  Right> Angled  Triangles. 

FIG.  18. 


Right-angled  triangles  (see  Fig.  18) 
may  be  solved  by  the  following  for- 
A      mulas : 


Solving  for  Any  Side. 

A  =  CXsmea      =  *X  tang.  *=—£.—  =  -*— 


B  =  C  X  cos.  a    =  A  X  cot.  a    =  - 


A 


sec. 


tang.  # 


c  = 


Solving  for  Any  Function  or  for  Any  Angle. 


Sin. 

a 

A 

—    c 

Tang. 

a 

A 
—    B 

Cos. 

a 

B 

—    c 

Cot.  a 

B 

~    A 

Sin. 

a 

=  cos.  b 

Tang. 

a 

=  cot. 

b 

Sin. 

b 

=  cos.  a 

Tang. 

b 

=  cot. 

a 

Sin. 

b 

B 

—  ~<r 

Tang. 

b 

B 

—     A 

Cos. 

b 

A 

—    c 

CotJ 

A 
—     B 

Angle  a  —  90°  —  b. 


Sec.  a    =  -Q- 

£ 

Cosec.  a=  —^ 

Sec.  a     =  cosec.  b 
Sec.  b     =  cosec.  a 

Sec.*     =4 
Cosec.  b  =  —- 


Angle  b  —  90°  —  a. 


Solving  for  Area. 

A   X  B 
Area  = 3— 

X  sin.  a  X  cos.  a        C2  X  cos.  b  X  sin.  b 


2 
X  tang,  a 


A  2  X  tang,  b 


TRIGONOMETRY. 


'77 


Fie.  1 9 


EXAMPLE. 

Find  angles  a  and  b  and  the 
side  X  in  the  right-angled  triangle. 
(Fig.  19). 


i»- 12.5  feet « 


Tangent  corresponding  to  a  =  ~f2jj-  =  0.4 

Tangent  corresponding  to  b  =  — j —  =  2.5 

By  the  trigonometrical  table  the  angles  are  obtained  thus : 

Tangent  0.40000  gives  21°  48' 

Tangent  2.50000  gives  68°  12' 
Therefore : 

Angle  a  =  21°  48'  and  angle  b  =  68°  12'. 
Angle  b  may  also  be  found  by  subtracting  angle  a  from 
90°,  thus : 

Angle  b  =  90°  —  21°  48'  =  68°  12' 
The  length  of  the  side  X  may  be  found  thus : 


X  — 


thus  : 


sin.  c 

5 
x  —   0.37137 

x  =  13.464  feet  long. 
By  means  of  logarithms  the  length  of  the  side  x  is  obtained 


Log.  x  —  log.  5  —  log.  sin.  21°  48' 
Log.  x  —  0.698970  —  (9.569804  —  10) 
Log.x—  1.129166 

x  —  13.464  feet  long. 

Solution  of  Oblique-Angled  Triangles. 

FIG.  20. 

/  Oblique-angled  triangles 

A      (see  Figs.  20-21-22)  may  be  solved 


(see  Y  igs.  20-21-22)  may  be  s 
by  the  following  formulas : 


178  TRIGONOMETRY. 

Solving  for  Any  Side. 


Csin.  b A  sin.  b 

sin.  c    ~  ~     sin.  a 


Solving  for  Any  Angle. 

Cos.  a  —  B*  "*"  C*  ~  A* 


2  A  C 


2A  B 

A  A 

Sin.  a  =  sin.  b  -  =  sin.  c  — 
B  C 

T>  ry 

Sin.  b  =  sin.  £  -  =  sin.  a  •—— 
C  A 

C  C 

Sin.  c  =  sin.  #  —  =  sin.  b  — 
A  B 

#  =  180°  —  (b  +  c) 
b  =  180°  —  (a  +  c) 
c  =180°  —  (a+b) 

Solving  for  Area. 

Area  —  sin-  c  X  A  x  B  —  sin,  a  X  C  X  ^?  _  sin,  b  X  A  X  C 

222 
EXAMPLE  1. 

Find  the  length  of  the  side  C  (see  Fig.  20)  when  angle  a  — 

20°  38'  12",  angle  c  —  117°  48'  5",  and  side  A  —  12.75  feet  long. 

NOTE.  —  The  angle  c  exceeds  90°,  therefore  the  supplement 

of  the  angle  must  be  used,  which   is  180°  —  117°  48'  5"  — 

62°  11'  55". 

Thus  the  solution  : 

c_  12.75  X  sin.  62°  11'  55" 

sin.  20°  38'  12" 
c_  12.75  X  0.88456 

0.35243 
C—  32  feet  long. 


TRIGONOMETRY.  179 

EXAMPLE  2. 

Find  the  length  of  the  side  B  (see  Fig.  20)  when  angle  b 
is  41°  33'  43",  side  C  is  32  feet  and  side  A  is  12.75  feet. 

In  this  example  two  sides  and  their  included  angle  are 
given  and  the  third  side  is  required  ;  therefore  the  formula 

B  —  \/A2  +  C2  —  2  A  C  cos.  b  must  be  used. 

Solution : 

B—  Vl2.752  +  322  —  2  X  12.75  X  32  X  0.748238 

B  —  VH86.562  —  610.562 

B  —  \/576  =  24  feet  long. 

EXAMPLE  3. 

Find  the  length  of  the  side  B  when  side  A  is  12.75  feet 
long,  angle  b  is  41°  33'  43"  and  angle  c  is  117°  48'  5' .  ( See 
Fig.  20). 

In  this  problem  one  side  and  its  two  adjacent  angles  are 
given ;  therefore  it  can  not  be  solved  directly  by  any  of  the 
preceding  formulas,  but  the  first  thing  to  do  is  to  find  the  angle 
opposite  to  side  A. 

Thus:  Angle  a  =  180°  —  (41°  33'  43"  +  117°  48'  5")  = 
20°  38'  12".  The  side  B  may  be  found  by  the  formula 

A  sin.  b 
**  sin.  a 

Solution : 

A  sin.  41°  33'  43" 


B  = 


sin.  20°  38'  12" 


B  =  12'75  *  0.66343 

0.35242 

B  =  24  feet  long. 
EXAMPLE  4. 

Find  length  of  the  side  C  when  B  is  24  feet  long,  angle  c  is 
117°  48'  5"  and  the  side  A  is  12.75  feet  long.     (  See  Fig.  20). 
Solution  : 


C  =      A*  +  J5*  —  2A£  cos.  c 

C  —  Vl2?752  —  242  —  2  X  12.75  X  24  X  (—  0.4664) 

C  —  Vl  62.56  +  576  +  285.44 

C  —  Vl02T  =  32  feet  long. 

NOTE.  —  In  this  example  the  cos.  of  117°  48'  5"  is  used, 
which,  in  numerical  value,  is  equal  to  cos.  of  62°  11'  55"  = 
0.4004,  but  cos.  in  the  second  quadrant  is  negative  (see  page  154): 
therefore  cos.  117°  48''  5"  =  (  —  0.40045)  and  the  essential  sign 
of  the  last  product  after  it  is  multiplied  by  this  negative  cos. 
must  change  from  —  to  -f  •  (  See  Algebra,  page  63). 


I  So  TRIGONOMETRY. 

EXAMPLE  5. 

Find  the  length  of  the  side  A  when  C  is  32  feet  long,  angle 
a  is  20°  38'  12"  and  angle  c  is  117°  48'  15". 
NOTE.— Supplement  to  c  is  62°  11'  55". 

Solution : 

_  Csin.  a 


A  = 


sin.  c 
32  X  0.35242 


0.88456 
A  —  12.75  feet  long. 

In  this  example,  as  in  the  preceding  one,  we  use  the  sup- 
plement of  the  angle  in  obtaining  its  function,  but  here  it  has  no 
influence  on  the  signs  because  sin.  is  positive  as  well  in  the 
second  as  in  the  first  quadrant. 

EXAMPLE  6. 

Find  angle  a  in  Fig.  20,  when  A  is  12.75  feet,  B  is  24  feet 
and  C  is  32  feet. 

Solution  : 


— 

_  576+  1024  —  162.5625 
C°S-  a  ^  -1636- 

cos.  a  —  0.93583 
Angle  a  —  20°  38'  12" 

EXAMPLE  7. 

Find  angle  ft,  Fig.  20,  by  the  same  formula. 

Solution  : 

co,»  =  -  +  --* 

,  _  12.752  +  322  —  242 
2  X  12.75  X  32 

cos,  ft  =  610'5625 
816 

cos.  ft  =  0.748238 
Angle  ft  =  41°  33'  43" 


TRIGONOMETRY.  l8l 


EXAMPLE  8. 

Find  angle  c,  Fig.  20,  by  the  same  formula. 

Solution : 


cos  e  _  12.752  +  242  — 322 


COS.  C  = 


2  X  12.75  X  4 
738.5625  —  1024 


612 

cos.  c  =  —  0.46640 
Supplement  to  angle  c  =  62°  11'  55",  and  angle  c  =  117°  48'  5" 

NOTE. — The  negative  cosine  indicates  that  it  is  in  the  sec- 
ond quadrant,  therefore  the  angle  is  over  90°. 

The  angle  corresponding  to  this  cosine  is  the  supplement 
of  angle  c.  To  obtain  angle  c,  the  angle  of  its  supplement  must 
be  subtracted  from  180°. 

EXAMPLE  9. 

Find  angles  a,  b  and  c  in  Fig.  20,  when  side  A  is  12.75 
feet,  B  32  feet,  and  C  24  feet. 


cos.^32^242-12-75" 


cos.  a  = 


2  X  32  X  24 
1437.4375 


1536 

cos.  a  =  0.93583 
Angle  a  =  20°  38'  12" 

Angle  b  may  be  found  by  the  formula  : 

n 

sin.  b  =  sin.  a  - 
A 

sin.  b  =  sin.  20°  38'  12"    — 
A 

sin.  b  =  0.35244  X  -    32 
12.75 

sin.  b  =  0.35244  X  1.8824 
sin.  b  =  0.66343 
Angle  b  =  41°  33'  43" 


1  82  TRIGONOMETRY. 

Angle  c  may  be  found  by  the  formula: 
c  =  180°  —  (a  +  b) 
c  =  180°  —  (20°  38'  12"  +  41°  33'  43") 
c  =  180°  —  62°  11'  55" 
c  =  117°  48'  5" 
EXAMPLE  10. 

Find  the  area  of  a  triangle  (see  Fig.  20),  when  it  is  known 
that  side  A  is  12.75  feet,  side  B  is  24  feet,  and  the  including 
angle  C=  117°  48'  5". 
Solution  : 

Sin.  to  supplement  of  117°  48'  5"  =  sin.  62°  11'  55"  = 
0.88456. 

Area  =  gjn.  C  X  ^  X  Z? 


Area  =  0-88456X^12.75X24  =  135  34  square  £cet 

EXAMPLE  11. 

Find  angle  c  and  the  sides  X  and  y  in  the  triangle,  Fig.  23. 

Solution  : 

c  =  180°  —  (40°  -f-  60°)  =  80°  F.Q.  23 

The  side  ;r=:?5_Xsm.40° 


•y  


sin.  60° 

25  X  0.64279 

0.86603 


x— 16-06975 

"  0.86603 
X  —  18.556  meters  long. 

By  the  use  of  logarithms  the  side  X  is  solved  thus : 
Log.  X  —  log.  25  -f  log.  sin.  40°  —  log.  sin.  60°. 
Log.X=  1.39794  +  (9.808067—10)  —  (9.937531—10). 
Log.X=  1.268476 

X  —  18.556  meters  long. 

The  side  y  =  25  X  sin'G80° 
sin.  60° 

_  25  X  0.98481 

0.86603 

y  —  28.429  meters  long. 

By  the  use  of  logarithms  the  sidejj/  is  solved  thus: 
Log.  y  =  log.  25  -f  log.  sin.  80°  —  log.  sin.  60°. 
Log.  y  —  1.39794  +  (9.993351—10)  —  (9.937531—10). 
Log.  y  =•  1.45376 

y  —  28.429  meters  long. 


TRIGONOMETRY.  183 


EXAMPLE  12. 

Find  angles  c  and  b  and  the  length 
of  the  side  X  in  Fig.  24. 


35 

Sin  c  —  42  X  0-80902 
35 

Sin.  c  =  0.97082  ^ ^^ „ 

Angle  c  =  76°  7'  26" 

Angle  £  =  180°  —  (54°  0'  0"  +  76°  T  26")  =  49°  52'  34" 

c. ,     y      35  X  sin.  49°  52'  34" 
bide  X  = — — 


=  ^^ 


0.80901 


By  means  of  logarithms  the  side  ^T  is  solved  thus  : 
Log.  X  —  log.  35  -f  /^.  sin.  49°  52'  34"  —  log.  sin.  54°. 
Log.  X^  1.544068  +  (9.883463—10)  —  (9.907958—10). 
Log.X—  1.519573 

X  —  33.08  meters  long. 

NOTE.  —  The  angle  c  is  obtained  by  interpolation  thus  :  In 
the  table  of  trigonometrical  functions  the  sine  0.97100  corre- 
sponds to  the  angle  76°  10'  and  the  sine  0.97030  corresponds  to 
the  angle  76°.  Thus,  a  difference  of  0.00070  in  the  sine  gives 
a  difference  of  10'  —  600"  in  the  angle. 

The  sine  to  angle  c  is  0.97082 
The  nearest  less  sine  in  the  table  is  0.97030  corresponding 

to  angle  76°  0'  0".  Difference,  0.00052 

Therefore  when  an  increase  in  sine  of  0.00070  corresponds 

to  an  increase  of  600"  in  the  angle,  an  increase  of  0.00052  will 

increase  the  angle  60°   X  °-00052==  446"  =  Q°  7'  26" 
0.00070 

thus,  the   angle  corresponding  to  the  sine    0.97082    must   be 
76°  7'  26". 


184 


GEOMETRY. 


PROBLEMS  IN  GEOMETRICAL  DRAWING. 


cf 

/,  e,  d, 


To  divide  a  straight  line  into  a 
given  number  of  equal  parts.  (See 
Fig.  1). 

Given  line  a  b,  which  is  to  be 
divided  into  a  given  number  of  equal 
parts.  Draw  the  line  b  c,  of  indefinite 
length,  and  point  off  from  b  the  re- 
quired number  of  equal  parts,  as  h,g, 
join  c1  and  0,  and  draw  the  other  lines  parallel  to  c1  a. 

To  erect  a  perpendicular  at  a 
given  point  on  a    straight   line. 


FIG.    1 


FIG.    2 


(See  Fig.  2). 

Given  line  a  b,  and  the  point 
x.  The  required  perpendicular 
is  x  y. 

5  Solution : 

'1  '2  With  x  as  center   and  any 

radius,  as  x  1,  cut  the  line  a  b  at  1  and  2.  With  1  and  2  as 
centers  and  with  a  radius  somewhat  greater  than  1  to  x,  describe 
arcs  intersecting  each  other  at  y.  Draw  x  y.  This  will  be  the 
required  perpendicular. 

From  a  given  point  without  a  straight  line  to  draw  a  per- 
pendicular to  the  line.  (See  Fig.  3). 

Given  line  a  b  and  the  point  c. 
The  required  perpendicular  is  x. 

Solution : 

With  the  point  c  as  center  and  any 
radius  as  c  1,  strike  the  arc  1  to  2.  With 
1  and  2  as  centers  and  any  suitable 
radius,  describe  arcs  intersecting  each 
other  at  «,  lay  the  straight  edge  through 
points  n  and  c  and  draw  the  perpendicular  x. 

To  erect  a  perpendicular  at  the  extremity  of  a  straight  line. 


FIG. 


(See  Fig.  4). 
li 


n 


The  required  perpendicular  is  x. 

Solution : 

From  any  point,  as  c,  with  radius 
as  a  c,  draw  the  circle.  From  point 
of  intersection,  #,  through  center,  c, 
draw  the  diameter  n  p.  From  the 
points,  through  the  point  of  intersec- 
tion at  p,  draw  the  perpendicular  x. 
__,  The  correctness  of  this  con- 

struction is  founded  on  the  principle 
that  inside  a  half  circle  no  other 


PROBLEMS  IN  GEOMETRICAL    DRAWING. 


'85 


FIG 


angle  but  an  angle  of  90°  can  simultaneously  touch  three  points 
in  the  circumference  when  two  of  these  points  are  in  the  point 
of  intersection  with  the  diameter  and 
the  circumference  and  the  third  one 
anywhere  on  the  circumference  of  the 
half  circle.  The  pattern  maker  is  mak- 
ing practical  use  of  this  geometrical 
principle,  when  he  by  a  common  car- 
penter's square  is  trying  the  correctness 
of  a  semi-circular  core  box,  as  shown 
in  Fig.  5. 


FIG.   6 


FIG.  7 


Draw  a  line  parallel  to   a 
given  line.     (See  Fig.  6). 

Given  line  a  b.     The  required 
line  x  y. 

Solution  : 

Describe  with  the  compass     a — 
from  the  line  a  b,  the  arcs  1  and 
2  ;  draw  line  x  y,  touching  these  arcs. 

To  divide  a  given  angle  into  two 
equal  angles. 

The  given  angle,  a  b  c,  is  divid- 
ed by  the  line  b  d. 

Solution : 

With  b  as  center  and  any  radius, 
as  b  1,  describe  the  arc  1  to  2.  With 
1  and  2  as  centers  and  any  suitable 
radius,  describe  arcs  cutting  each 
other  at  d.  Draw  line  b  d,  which 
will  divide  the  angle  into  two  equal  parts. 


To  draw  an  angle  equal  to  a  given  angle. 

Given  angle  a  b  c.  Construct 
angle  x  y  z. 

With  b  as  center  and  any  radius, 
as  b  1,  describe  the  arc  1  to  2. 
Using  y  as  center  and  without  alter- 
ing the  compass,  describe  the  arc  /, 
intersecting  y  z.  Measuring  the 
distance  from  2  to  1  on  the  given 
angle,  transfer  this  measure  to  the  2 

arc  /,  through  the  point  of  intersection.     Draw  the  line  y  *•, 
and  this  angle  will  be  equal  to  the  first  angle. 

NOTE. — Angles  are  usually  measured  by  a  tool  called  a  pro- 
tractor, looking  somewhat  like  Fig.  9  or  10,  usually  made  from 
metal,  and  supplied  by  dealers  in  draughting  instruments.  A 


FIG.  8 


i86 


PROBLEMS    IN    GEOMETRICAL   DRAWING. 


protractor  may  also  be  constructed  on  paper  and  used  for 
measuring  angles,  but  it  should  then  always  be  made  on  as 
large  a  scale  as  convenient. 


FIG.    9 


FIG.  10 


To  draw  a  protractor  with  a  division  of  5°.    (See  Fig.  10). 

Construct  an  angle  of  exactly  90  degrees,  divide  the  arc  into 
nine  equal  parts,  then  each  part  is  10C;  divide  each  part  into 
two  equal  parts  and  each  is  5°. 


Prove  that  the  sum  of  the  three  angles  in  a  triangle  consists 

of  180°.     (See  Fig.  11). 
a        F|C-  11  Solution: 

In  the  triangle  a  b  c,  extend 
the  base  line  to  i.  Draw  the  line 
o  p,  parallel  to  the  side  a  b, 
thereby  the  angle  g  will  be  equal 
to  the  angle  d,  and  the  angle  h 
must  be  equal  to  angle  c.  The 
angle/is  one  angle  in  the  triangle 
and  /+  g  -\-  h  =  180°,  therefore 
/  +  d  +  c  must  also  be  180°. 


To  draw  on  a  given  base  line  a  triangle  having  angles  90°, 
30°  and  60°.    (See  Fig.  12). 

Given  line  a  b,  required  triangle  is  a,  c,  b. 
Solution : 

Extend  the  line  a  b  to  twice  its 
length,  to  the  point  e.  With  e  and  b  as 
centers  strike  arcs  intersecting  each 
other  and  erect  the  perpendicular  a  c. 
With  b  as  center  and  any  radius  as  /, 
draw  the  arc  /  m.  With  /  as  center 
and  with  the  same  radius,  describe  arc 
intersecting  at  m.  From  b  through 
point  of  intersection  at  m,  draw  line  b 
intersecting  the  perpendicular  at  c. 
This  will  complete  the  triangle. 


FIG.  12 


PROBLEMS  IN  GEOMETRICAL  DRAWING. 


l87 


To  draw  a  square  inside  a 
given  circle.  (See  Fig.  13). 

Solution : 

Draw  the  line  a  b  through  the 
center  of  the  circle.  From  points 
of  intersection  at  a  and  b,  describe 
with  any  suitable  radius  arcs  inter- 
secting at;*  and  ///.  Draw  through 
the  points  the  line  c  d.  Connect 
the  points  of  intersection  on  the 
circle  and  the  required  square  is 
constructed. 


FIG.  13, 


To  draw  a  square  outside  a 
given  circle.  (See  Fig.  14). 

Solution : 

Draw  lines  a  b  and  c  d,  and 
from  points  of  intersection  at  b  and 
r,  describe  half  circles;  their  points 
of  intersection  determine  the  sides 
of  the  square. 


To  draw  a  hexagon  within  a  given 
circle.  (See  Fig.  15). 

Apply  the  radius  as  a  chord  succes- 
sively about  the  circle;  the  resulting 
figure  will  be  a  hexagon. 


FIG.  15- 


FIG.  16. 


To  inscribe  in  a  circle  a  regular  polygon  of 
any  given  number  of  sides. 

Solution: 

Divide  .360  by  the  number  of  sides,  and  the 
quotient  is  the  number  of  degrees,  minutes,  and 
seconds  contained  in  the  center  angle  of  a  triangle, 
of  which  one  side  will  make  one  of  the  sides  in 
the  polygon.  For  instance,  draw  a  hexagon  by  this  method. 
(See  Fig.  10).  360  =  6QO 

6 


i88 


PROBLEMS    IN    GEOMETRICAL    DRAWING. 


FIG.   17 


To  find  the  center  in  a  given 
circle.  (See  Fig.  17). 

Solution : 

Draw  anywhere  on  the  circumfer- 
ence of  the  circle  two  chords  at  ap- 
proximately right  angles  to  each  other, 
bisect  these  by  the  perpendiculars  x 
and  ^,  and  their  point  of  intersection 
is  the  center  of  the  circle. 


FIG.    18 


To    draw    any  number  of 
circles    between    two    inclined 
lines  touching  themselves  and 
the  lines.     (See  Fig.  18). 
Solution : 

Draw  center  line  ef.  Draw 
first  circle  on  line  i  g.  From 
point  of  intersection  between 
this  circle  and  the  center  line  draw  the  line  //,  perpendicular  to 
a  b.  Describe  with  a  radius  equal  to  ^,  the  arc  intersecting  at 
g\  draw  line^1 11,  parallel  to  g  i,  and  its  point  of  intersection 
with  the  center  line  gives  the  center  for  the  next  circle,  etc. 


FIG. 


-Xr-4 1- 


To  draw  a  circle  through 
three  given  points.  (See  Fig.  19). 

The  given  points  are  #,  £,  and  c. 

Solution : 

From  a  and  b  as  centers  with 
suitable  radius,  describe  arcs  inter- 
secting at  e  e.  Draw  a  line  through 
these  points.  From  b  and  c  as  cen- 
ters, describe  arcs  intersecting  at  d 
d;  draw  a  line  through  these  points. 
The  point  where  these  two  lines  intersect  is  the  center  of  the 
circle. 


FIG.  20 


To  draw  two  tangents  to  a 
circle  from  a  given  point  without 
same  circle.  (See  Fig.  20). 

Given  point  «,  and  the  circle 
with  the  center  «.  The  required 
tangents  are  a  d,  and  a  b. 

Solution : 

Bisect  line  n  a.  With  c  as 
center  and  radius  a  c,  describe 


X^SE 

;  NIVERSF 


PROBLEMS  IN  GEOMETRICAL  DRAWING. 


the  arc  b  d  through  the  center  of  the  circle.  The  points  of 
intersection  at  b  and  d  are  the  points  where  the  required  tan- 
gents a  b  and  a  d  will  touch  the  circle. 


F/G 


FIG.  22. 


To  draw  a  tangent  to  a  given  point  in 
a  given  circle.  (See  Fig.  21). 

Given  circle  and  the  point  h,  x  y  is 
required. 

Solution : 

The  radius  is  drawn  to  the  point  h  and 
a  line  constructed  perpendicular  to  it  at  the 
point  h.  This  perpendicular,  touching  the 
circle  at  ^,  is  called  a  tangent. 

To  draw  a  circle  of  a  certain  size  that 
will  touch  the  perphery  of  two  given  cir- 
cles. (See  Fig.  22). 

Given  the  diameter  of  circles  a,  b, 
and  c.  Locate  the  center  for  circle  £, 
when  centers  for  a  and  b  are  given. 

Solution : 

From  center  of  a,  describe  an  arc 
with  a  radius  equal  to  the  sum  of  radii  of  a  and  c.  From  b  as 
center,  describe  another  arc  using  a  radius  equal  to  the  sum  of 
the  radii  of  b  and  c.  The  point  of  intersection  of  those  two 
arcs  is  the  center  of  the  circle  c. 

NOTE. — This  construction  is  useful  when  locating  the  center 
for  an  intermediate  gear.  For  instance,  if  a  and  b  are  the  pitch 
circles  of  two  gears,  c  would  be  the  pitch  circle  located  in  correct 
position  to  connect  a  and  b. 

To  draw  an  ellipse,  the  longest 
and  shortest  diameter  being  given. 
The  diameters  a  b  and  c  d  are 
given.  The  required  ellipse  is 
constructed  thus :  (See  Fig.  23). 

From  c  as  center  with  a  radius 
a  n,  describe  an  arc  fl  f.  The 
points  where  this  arc  intersects  a 
b  are  foci.  The  distance  fn  is 
divided  into  any  number  of  parts, 
as  1,  2,  3,  4,  5.  With  radius  1  to  b, 
and  the  focus/ as  center,  describe  arcs  6  and  61 ;  with  the  same 
radius  and  with  f1  as  center  describe  arcs  62  and  63.  With 
radius  1  to  a  and/1  as  center,  describe  arcs  intersecting  at  6  and 
61 ;  with  the  same  radius  and  with  /  as  center,  describe  arcs 
intersecting  at  62  and  68.  Continue  this  operation  for  points  2, 
3,  etc.,  and  when  all  the  points  for  the  circumference  are  in  this 


FIG.    23 


190 


PROBLEMS    IN    GEOMETRICAL    DRAWING. 


way  marked  out,  draw  the  ellipse  by  using  a  scroll.  It  is  a 
property  with  ellipses  that  the  sum  of  any  two  lines  drawn  from 
the  foci  to  any  point  in  the  circumference  is  equal  to  the  largest 
diameter.  For  instance  : 


e  +  fe,  =  ab,  or/61 


61,  =  a  b. 


Cycloids. 

Suppose  that  a  round  disc,  c,  rolls  on  a  straight  line,  a  b,  and 
that  a  lead  pencil  is  fastened  at  the  point  r;  it  will  then  describe 

a  curved  line,  a,  /,  r,  ;/,  b.  This 
line  is  called  a  cycloid.  (See 
Fig.  24). 

This  supposed  disk  is  usual- 
ly called  the  generating  circle. 
The  line  a  b  is  the  base  line  of 
the  cycloid  and  is  equal  in  length 

to  TT  times  m  r,  or  practically  3.1416  times  the  diameter  of  the 
generating  circle.  The  length  of  the  curved  line  a,  /,  r,  «,  3,  is 
four  times  r  m,  (four  times  as  long  as  the  diameter  of  the 
generating  circle). 

A  circle  rolling  on  a  straight  line  generates   a  cycloid. 
(See  Figs.  24  and  25). 

A  circle  rolling    upon    another    circle  is    generating    an 
epicycloid.    (See  Fig.  26). 

A  circle  rolling  within  another  circle  generates  a  hypo- 
cycloid.     (See  Fig.  27). 

To  draw  a  cycloid,  the  generating  circle  being  given. 

Solution : 

Divide  the  diameter  of  the 
rolling  circle  in  7  equal  parts. 
Set  off  11  of  these  parts  on  each 
side  of  a  on  the  line  d  e.  This 
will  give  a  base  line  practically 
equal  to  the  circumference. 
Divide  the  base  line  from  the 

point  a  into  any  number  of  equal  parts;  erect  the  perpendicu- 
lars, with  center-line  as  centers  and  a  radius  equal  to  the  radius 
of  the  generating  circle  describe  the  arcs.  On  the  first  arc  from 
d  or  e  set  off  one  part  of  the  base  line.  On  the  second  arc  set 
off  two  parts  of  the  baseline;  on  the  third  arc  three  parts,  etc. 
This  will  give  the  points  through  which  to  draw  the  cycloid. 


PROBLEMS  IN  GEOMETRICAL    DRAWING. 


191 


To  draw  an  epicycloid  (see  Fig.  26),  the  generating  circle  a 
and  the  fundamental  circle  B  being  given. 


Solution 


FIG.  26 


Concentric  with  the  circle  B,  describe 
an  arc  through  the  center  of  the  generating 
circle.  Divide  the  circumference  of  the 
generating  circle  into  any  number  of  equal 
parts  and  set  this  off  on  the  circumference 
of  the  circle  B.  Through  those  points  draw 
radial  lines  extending  until  they  intersect 
the  arc  passing  through  the  center  of  the 
generating  circle.  These  points  of  inter- 
section give  the  centers  for  the  different  positions  of  the  gener- 
ating circle,  and  for  the  rest,  the  construction  is  essentially  the 
same  as  the  cycloids.  In  Fig.  26,  the  generating  circle  is  shown 
in  seven  different  positions,  and  the  point  «,  in  the  circumfer- 
ence of  the  generating  circle,  may  be  followed  from  the  position 
at  the  extreme  left  for  one  full  rotation,  to  the  position  where  it 
again  touches  the  circle  B. 


To  draw  a  hypocycloid.     (See  Fig.  27). 

The  hypocycloid  is  the  line  generated 
by  a  point  in  a  circle  rolling  within  another 
larger  circle,  and  is  constructed  thus:  (See 
Fig.  27). 


FIG-  27. 


Divide  the  circumference  of  the  gener- 
ating circle  into  any  number  of  equal  parts. 
Set  off  these  on  the  circumference  of  the 
fundamental  circle.  From  each  point  of 
division  draw  radial  lines,  1,  2,  3,  4,  5,  6. 
From  11  as  center  describe  an  arc  through 
the  center  of  the  generating  circle,  as  the 
arc  c  d.  The  point  of  intersection  between  this  arc  and  the 
radial  lines  are  centers  for  the  different  positions  of  the  gener- 
ating circle.  The  distance  from  1  to  a  on  the  fundamental 
circle1  is  set  off  from  1  on  the  generating  circle  in  its  first  new 
position ;  the  distance  2  to  a  on  the  fundamental  circle  is  set  off 
from  2  on  the  generating  circle  in  its  second  position,  etc.  For 
the  rest,  the  construction  is  substantially  the  same  as  Figs.  25 
and  26. 


NOTE. — If  the  diameter  of  the  generating  circle  is  equal  to 
the  radius  of  the  fundamental  circle,  the  hypocycloids  will  be  a 
straight  line,  which  is  the  diameter  of  the  fundamental  circle. 


192 


PROBLEMS    IN    GEOMETRICAL   DRAWING. 


Involute. 

An  involute  is  a  curved  line  which  may  be  assumed  to  be 
generated  in  the  following  manner  :    Suppose  a  string  be  placed 
around  a  cylinder  from  a  to  ^,  in  the 
direction  of  the  arrow  (see  Fig.  28), 
and  having  a  pencil  attached  at  b ; 


FIG.  28 


keep  the  string  tight  and  move  the 
pencil  toward  r,  and  the 


involute, 
b  c,  is  generated. 

To  draw  an  involute. 
Solution : 

From  the  point  b,  (see  Fig.  28) 
set  off  any  number  of  radial  lines  at 
equal  distances,  as  1,  2,  3, 4, 5.  From 
points  of  intersection  draw  the  tangents  (perpendicular  to  the 
radial  lines).  Set  off  on  the  first  tangent  the  length  of  the  arc  1 
to  b\  on  the  second  tangent  the  arc  2  to  b,  etc.  This  will  give 
the  points  through  which  to  draw  the  involute. 

To  draw  a  spiral  from  a  given 
F,G.   29  point,*. 

Solution: 

Draw  the  line  a  b  through  the 
point  c.  Set  off  the  centers  r  and 
S,  one-fourth  as  far  from  c  as  the 
distance  is  to  be  between  two  lines 
in  the  spiral.  Using  r  as  center, 
describe  the  arc  from  c  to  1,  and 
using  ,5*  as  center,  describe  the  arc 
from  1  to  2  ;  using  r  as  center,  de- 
scribe the  arc  from  2  to  3,  etc. 


FIG. 


Conical  Sections. 

If  a  cone  (see  Fig.  30),  is  cut  by  a  plane  on  the  line  a  b, 
which  is  parallel  to  the  center  line,  the 
section  will  be  a  hyperbola, 

If  cut  by  a  plane  on  the  line  c  d,  which 
is  parallel  to  the  side,  the  section  will  be  a 
parabola. 

If  cut  by  a  plane  on  the  line^,  which 
is  parallel  to  the  base  line,  the  section  will 
be  a  circle. 

If  cut  by  a  line,  e  /,  which  is 
neither  parallel  to  the  side,  the  center- 
line  nor  the  base,  the  section  will  be  an 
ellipse* 


MENSURATION.  193 

MENSURATION. 

If  each  side  in  a  square  (see  Fig.  1)  is  two  feet  long,  the 
area  of  the  figure  will  be  4  square  feet;   that  is,  it  contains  four 
squares,  each  of  which  is  one  square  foot. 
Thus  the  area  of  any  square  or  rectangle  is  FIG.  1 

calculated  by  multiplying  the  length  by  the 
width. 

EXAMPLE  1. 

What  is  the  area  of  a  piece  of  land 
having  right  angles  and  measuring  108  feet 
long  and  20  feet  wide  ? 

Solution : 

108  X  20  =  2160  square  feet. 


-2  :feet 


EXAMPLE  2. 

What  is  the  area  in  square  meters  of  a  square  house-lot  30 
meters  long  and  30  meters  wide. 
Solution : 

30  X  30  =  900  square  meters. 

(  Square  meter  is  frequently  written  m*  and  cubic  meter  is 
written  ms). 

A  square  inscribed  in  a  circle  is  half  in  area  of  a  square 
outside  the  same  circle.  Divide  the  side  of  a  square  by  0.8862, 
and  the  quotient  is  the  diameter  of  a  circle  of  the  same  area  as 
the  square. 

The  Difference  between  One  Square  Foot  and  One  Foot 
Square. 

One  foot  square  means  one  foot  long  and  one  foot  wide, 
but  one  square  foot  may  be  any  shape,  providing  the  area  is  one 
square  foot.  For  instance,  Fig.  1  is  two  feet  square,  but  it  con- 
tains four  square  feet.  One  inch  square  means  one  inch  long 
and  one  inch  wide,  but  one  square  inch  may  be  any  shape, 
provided  the  area  is  one  square  inch.  One  mile  square  means 
one  mile  long  and  one  mile  wide,  but  one  square  mile  may  have 
any  shape,  provided  the  area  is  one  square  mile. 


Area  of  Triangles. 

The  area  of  any  triangle  may  be  found  by  multiplying  the 
base  by  the  perpendicular  height  and  dividing  the  product  by  2. 


194  MENSURATION. 

EXAMPLE. 

Find  the  area  of  a  triangle  16  inches  long  and  5  inches  per- 
pendicular height. 
Solution : 

Area  =  5  X  16  =  40  square  inches. 

The  perpendicular  height  in  any  triangle  is  equal  to  the 
area  multiplied  by  2  and  the  product  divided  by  the  base. 

The  area  of  any  triangle  is  equal  to  half  the  base  multiplied 
by  the  perpendicular  height. 

The  perpendicular  height  of  any  equilateral  triangle  is 
equal  to  one  of  its  sides  multiplied  by  0.866. 

The  area  of  any  equilateral  triangle  may  be  found  by  mul- 
tiplying the  square  of  one  of  the  sides  by  0.433. 

EXAMPLE. 

f  Find  the  area  of  an  equilateral  triangle  when  the  sides  are 
12  inches  long. 
Solution  : 

Area  =  12  X  12  X  0.433  =  62.352 

The  side  of  any  equilateral  triangle  multiplied  by  0.6582 
gives  the  side  of  a  square  of  the  same  area. 

The  side  of  any  equilateral  triangle  divided  by  1.3468  gives 
the  diameter  of  a  circle  of  the  same  area. 


To  Figure  the  Area  of  Any  Triangle  when  Only  the 
Length  of  the  Three  Sides  is  Given. 

RULE. 

From  half  the  sum  of  the  three  sides  subtract  each  side 
separately;  multiply  these  three  remainders  with  each  other 
and  the  product  by  half  the  sum  of  the  sides,  and  the  square 
root  of  this  result  is  the  area  of  the  triangle. 

EXAMPLE. 

Find  the  area  of  a  triangle  having  sides  12  inches,  9  inches 
and  15  inches  long. 

Solution : 

Half  the  sum  of  the  sides  =  18 

Area  =  V(18  — 12)  X  (18  —  9)  X  (18  —  15)  X  18 

Area  =  \/6  X  9  X  3  X  18 

Area  =  \/2916 

Area  =  54  square  inches. 


MENSURATION.  195 

To  Find  the  Height  in  any  Triangle  when  the   Length 
of  the  Three  Sides  is  Given. 

(See  Fig.  2). 

The  base  line  is  to  the  sum  of  the  other 
two  sides  as  the  difference  of  the  sides  is  to 
the  difference  between  the.  two  parts  of  the 
base  line,  on  each  side  of  the  line  measuring 

the  perpendicular  height.      If  half  this  dif-     

ference   is   either    added   to    or   subtracted  c 

from  half  the  base  line,  there  will  be  obtained  two  right-angled 
triangles,  in  which  the  base  and  hypothenuse  are  known  and 
the  perpendicular  may  be  calculated  thus :  Using  Fig.  2  for  an 
example,  and  adding  half  the  difference  to  half  the  base  line, 
this  may  be  written  in  the  formula: 

-*/~^2        f  (a  4-  b}  X  (a  —  b} \      c  \2 

x  —  \  a*  —  f  - ^ :  +  I 

V  V  2c  2  / 


RULE. 

Multiply  the  sum  of  the  sides  by  their  difference  and  divide 
this  product  by  twice  the  base;  to  the  quotient  add  half  the 
base;  square  this  sum  (that  is,  multiply  it  by  itself);  subtract 
this  from  the  square  of  the  longest  side,  and  the  square  root  of 
the  difference  is  the  perpendicular  height  of  the  triangle. 

EXAMPLE. 

In  the  triangle,  Fig.  2,  the  sides  are: 

c  =  12  inches. 

a  =  9  inches. 

b  =  6  inches  ;   find  the  perpendicular  height  x. 


_  /  (9  +  6)  X  (9  -  6)        12_ 
2  X  12  2 


x—  ~    81  —  7.875 2 


x-  =  J  81  —  62.015 
x=  J  18.985 
x  =  4.357  inches. 


196  MENSURATION. 

To  Find  the  Area  of  a  Parallelogram. 

Multiply  the  length  by  the  width,  and  the  product  is  the 
area. 

NOTE. — The  width  must  not  be  measured  on  the  slant  side, 
but  perpendicular  to  its  length. 

To  Find  the  Area  of  a  Trapezoid. 

Add  the  two  parallel  sides  and  divide  by  two  ;  multiply  the 
quotient  by  the  width,  and  the  product  is  the  area.  (See  Fig.  3). 

FIG.  3.  EXAMPLE. 

h  7  feet.  -^  F.nd    the    area    of     a     trapezojd 

(Fig.  3). 

Solution : 


Area  =  '  "*"  v  X  4  —  32  square  feet. 


NOTE.  —  The  correctness  of  this  may  be  best  understood  by 
assuming  the  triangle  b  cut  off  and  placed  in  the  position  a, 
and  the  trapezoid  will  be  changed  into  a  rectangle  8  feet  long 
and  4  feet  wide. 

The  area  of  any  polygon  may  be  found  by  dividing  it  into 
triangles  and  calculating  the  area  of  each  separately,  and  the 
sum  of  the  areas  of  all  the  triangles  is  the  area  of  the  polygon. 

The  Area  of  a  Circle. 

The  area  of  a  circle  is  equal  to  the  square  of  the  radius 
multiplied  by  3.1416,  which  written  in  a  formula  is, 
Area  =  3.1416  r*. 

The  area  of  a  circle  is  also  equal  to  the 
FIG.  *.  square  of  the  diameter  multiplied  by  0.7854, 

which  may  be  written, 

Area  =  0.7854  </2 

The  area  of  a  circle  is  also  equal  to  its 
circumference  multiplied  by  the  radius  and 
the  product  divided  by  2,  which  may  be 
written, 

Area  =  C-*^ 


The  correctness  of  these  formulas  may  be  best  understood 
by  assuming  the  circle  to  be  divided  into  triangles  (see  Fig.  4),  of 
which  the  height  //  —  radius  and  the  sum  of  the  bases,  b,  of  all 
the  triangles  is  equal  to  the  circumference  of  the  circle. 


MENSURATION.  1 97 

Therefore,  according  to  the  formulas, 
the  area  of  a  triangle  =  base  X  perpendicular  height 

the  area  of  a  circle  must  be  =  radius  X  circumference 

2 
and  from  this  follow  all  the  other  formulas. 

To  Change  a  Circle  into  a  Square  of  the  Same  Area. 

RULE. 

Multiply  the  diameter  of  the  circle  by  the  constant  0.8862 
and  the  product  is  the  length  of  one  side  in  a  square  of  the  same 
area. 

EXAMPLE. 

A  circular  water-tank  5  feet  in  diameter  and  3  feet  high  is 
to  be  replaced  by  a  square  tank  of  the  same  height  and  volume. 
How  long  will  each  side  in  the  new  tank  be? 

Solution: 

Side  =  5  X  0.8862  =  4.431  feet  long. 

To  Find  the  Side  of  the  Largest  Square  which  can  be 
Inscribed  in  a  Circle. 

RULE. 

Multiply  the  diameter  of  the  circle  by  the  constant  0.7071 ; 
the  product  is  the  length  of  the  side  of  the  square. 

EXAMPLE. 

What  is  the  largest  square  beam  which  can  be  cut  from  a 
log  30  inches  in  diameter. 

Solution : 

30  X  0.7071  =  21.213  inches  square. 

NOTE. — A  round  log  of  any  diameter  will  always  cut  into  a 
square  beam  having  sides  seven-tenths  the  diameter  of  the 
round  log.  For  instance,  a  10-inch  log  will  cut  7  inches  square, 
a  15-inch  log  will  cut  10.5  inches  square,  a  20-inch  log  will  cut 
14  inches  square,  etc. 

To  Find  the  Area  of  Any  Irregular  Figure. 

(See  Fig.  5).  F|G- 

Divide  the  figure  into  any 
number  of  equal  parts,  as  shown 
by  the  perpendiculars  1,  2,  3,  etc. 
Measure  the  width  of  the  figure 
at  the  middle  of  each  division; 
add  these  measurements  together, 


MENSURATION. 


divide  this  sum  by  the  number  of  divisions  (in  Fig.  5  it  is  '  ), 
multiply  this  quotient  by  the  length  a  b,  and  the  product  is  the 
area,  approximately. 

NOTE.  —  Sometimes  the  figure  is  of  such  shape  that  it  is 
more  convenient  to  divide  some  of  it  into  squares,  rectangles, 
or  triangles,  and  figure  the  rest  as  explained  above. 


To  Find  the  Area  of  a  Sector  of  a  Circle. 

The  area  of  a  sector  of  a  circle  is  to  the  area  of  the  whole 
circle  as  the  number  of  degrees  in  the  arc  of  the  sector  is  to  300 
degrees. 

Thus: 


A  =  r*  X  816  X  *    =    0.008727 


r 
I  =  3-1416  X  a  X  r  =  0  01745329  x  a  x  r 

r=J  360  A     _    10.7046A/— 
^  3.1416  a  >  a 

a  =     180/     —  57.2956  / 

3714167-  "  r 

A  =  Area  of  sector. 
r  =  radius  of  sector. 
a  =  number  of  degrees  in  arc. 
/=  length  of  arc  in  same  units  as  A  and  r. 

EXAMPLE. 

The  arc  of  the  sector  (Fig.  6)  is  60°  no. 

and  the  radius  is  6  feet.     Find  area. 

360         TT  r2 


60         Area 

)  ?*2  7T 

860 

Area  =  60  X  6  X  6  X  3-1416 
36~0 

Area  =  18.849  square  feet. 


If  the  length  of  the  arc  is  known  instead  of  the  number  of 
degrees,  multiply  the  length  of  the  arc  by  the  length  of  the 
radius,  divide  product  by  2,  and  the  quotient  is  the  area  of  the 
sector.  The  correctness  of  this  rule  will  be  understood  by  the 
rule  for  area  of  circles,  explained  under  Fig.  4. 


MENSURATION. 


I99 


To  Find  the  Length  of  Arc  of  a  Segment  of  a  Circle. 

The  length  of  the  arc  may  be  calculated  by  the  formula,* 
7       8c—  C 


1-=  Length  of  arc,  a  fb  \ 

c  =  Length  of  chord  from  a  tof  (   (  See  Fig.  7). 

C  =  Length  of  chord  from  a  to  b  ) 

RULE. 

Multiply  the  length  of  the  chord  of  half  the  arc  by 
8  ;  from  the  product  subtract  the  length  of  the  chord  of  the  arc  ; 
divide  the  remainder  by  3,  and  the  quotient  is  the  length  of 
the  arc. 

When  chord  and  height  of  segment  are  known,  the  chord  of 
half  the  arc  is  calculated  thus  : 

Chord  of  half  the  arc  =  f^/  ni  _|_  ^2 

h  =  Height  of  segment  (see  d  f,  Fig.  7). 

n  =  Half  the  length  of  chord  (see  a  d  or  b  d,  Fig.  7). 

When  only  the  radius  and  the  height  of  the  segment  are 
known,  the  length  of  the  chord  of  the  whole  arc  expressed  in 
these  terms  will  be  :  2  X  *J  2  r  /t  _  /fc2 

The  chord  of  half  the  arc  will  be  : 


r  fr 


Therefore  the  length  of  the  arc  will  be : 
\/2r  h  —  2  ) 


I  = 


8  X 


r  h  — 


/=  length  of  arc  (afb,  Fig.  7). 
h  =  height  of  segment  ( df,  Fig.  7 ). 
r  —  radius  of  circle  (c  f,  Fig.  7). 

To  Find  the  Area  of  a  Segment  of  a  Circle. 

(See  Fig.  7). 

Ascertain  the  area  of  the  whole 
sector  and  from  this  area  subtract  the 
area  of  the  triangle,  and  the  rest  is  the 
area  of  the  segment. 

EXAMPLE. 

Find  the  the  area  of  the  segment 
when  the  radius  is  9  inches  and  the 
arc  60°. 

*  This  formula  is  called  "  Huyghens's  approximate  formula  for  circular  arcs," 
but  it  is  so  close  that  it  may  for  any  practical  purpose  be  considered  absolutely  cor- 
rect for  arcs  having  small  center  angles;  for  center  angles  as  large  as  120  ,  the 
result  is  only  one  quarter  of  one  per  cent,  too  small,  and  even  for  half  a  circle  the 
result  is  scarcely  more  than  one  per  cent,  small  as  compared  to  results  calculated  by 
taking  TT  as  3.1416. 


FIG.    7 


2OO  MENSURATION. 

Solution  : 

Area  .of  segment  =  A  —  ££*JL  —  0.433  r2 
360 

A  =  60X9X9X3.1416_  g      Q 

360 

A  —  42.4116  —  35.073 
A  =  7.3386  square  inches. 

In  this  example  the  arc  was  60°,  consequently  the  triangle 
is  equilateral  ;  therefore  its  area  is  found  by  the  formula 
0.433  r2.  (  See  area  of  equilateral  triangles,  page  194). 

NOTE.  —  When  the  segment  is  greater  than  a  semicircle, 
calculate  by  preceding  rules  and  formulas  the  area  of  the  lesser 
portion  of  the  circle  ;  subtract  it  from  the  area  of  the  whole 
circle.  The  remainder  is  the  area  of  the  segment. 


To  Find  the  Radius  Corresponding  to  the  Arc,  when  the 
Chord  and  the  Height  of  the  Segment  Are  Given. 

RULE. 

Add  the  square  of  the  height  to  the  square  of  half  the 
chord  ;  divide  this  sum  by  twice  the  height,  and  the  quotient  is 
the  radius.  In  a  formula  this  may  be  written  : 


I  —  radius  —  c  b  or  cf 


( See  Fig.  7). 


n  =  half  the  chord  =  db 
h  —  height  —  df 

The  above  rule  and  formula  may  be  proved  by  rules  for 
right-angled  triangles;  thus,  c  b  or  r  equals  hypothenuse,  and  ;/, 
or  half  the  chord,  equals  perpendicular,  and  c  d,  which  is  equal 
tor  —  h,  is  the  base.  From  the  rule  that  the  square  of  the 
hypothenuse  is  equal  to  the  sum  of  the  square  of  the  base  and 
the  square  of  the  perpendicular,  we  have  : 


2  rh  = 


The  perpendicular  height  of  the  triangle  is  always  equal  to 
the  radius  minus  the  height  of  the  segment.  (  See  triangle  a  b  c, 
and  height,  df,  Fig.  7). 


MENSURATION. 


201 


TABLE  No.  23 Areas  of  Segments  of  a  Circle. 

The  diameter  of  a  circle  =  1,    and  it  is  divided  into  100 
equal  parts. 


h 

D 

Area. 

k 

D 

Area. 

h 
D 

Area. 

0.01 

0.001329 

0.18 

0.096135 

0.35 

0.244980 

0.02 

0.003749 

0.19 

0.103900 

0.36 

0.254551 

0.03 

0.006866 

0.20 

0.111824 

0.37 

0.264179 

0.04 

0.010538 

0.21 

0.119898 

0.38 

0.273861 

0.05 

0.014681 

0.22 

0.128114 

0.39 

0.283593 

0.06 

0.019239 

0.23 

0.136465 

0.40 

0.293370 

0.07 

0.024168 

0.24 

0.144945 

0.41 

0.303187 

0.08 

0.029435 

0.25 

0.153546 

0.42 

0.313042 

0.09 

0.035012 

0.26 

0.162263 

0.43 

0.322928 

0.10 

0.040875 

0.27 

0.171090 

0.44 

0.332843 

0.11 

0.047006 

0.28 

0.180020 

0.45 

0.342783 

0.12 

0.053385 

0.29 

0.189048 

0.46 

0.352742 

0.13 

0.059999 

0.30 

0.198168 

0.47 

0.362717 

0.14 

0.066833 

0.31 

0.207376 

0.48 

0.372704 

0.15 

0.073875 

0.32 

0.216666 

0.49 

0.382700 

0.16 

0.081112 

0.33 

0.226034 

0.50 

0.392699 

0.17 

0.088536 

0.34 

0.235473 

Table  No.  23  gives  the  areas  of  segments  from  0.01  to  0.5 
in  height  when  the  diameter  of  the  circle  is  1. 

The  area  of  any  segment  is  computed  by  the  following 
rule: 

Divide  the  height  of  the  segment  by  the  diameter  of  its 
corresponding  circle.      Find  in   the  table  in  column  marked 

-jj-  the  number  which  is  nearest,  and  multiply  the  corresponding 
area  by  the  square  of  the  diameter  of  the  circle,  and  the 
product  is  the  area  of  the  segment. 

EXAMPLE. 

Figure  the  area  of  a  segment  of  a  circle,  the  height  of  the 
segment  being  12  inches  and  the  diameter  of  the  circle  40  inches. 


Solution : 


12  divided  by  40  =  0.3 


In  the  column  marked  —p  find  0.3;  the  corresponding 
area  is  0.198168. 

The  area  of  the  segment  is  40  X  40  X  0.198168  =  317.0688 
square  inches,  or  317  square  inches, 


202 


MENSURATION. 


To  Calculate  the  Number  of  Gallons  of  Oil  in  a  Tank. 

EXAMPLE. 

A  gasoline  tank  car  is  standing  on  a  horizontal  track,  and 
by  putting  a  stick  through  its  bung-hole  on  top  it  is  ascertained 
that  the  gasoline  stands  15  inches  high  in  the  tank.  The 
diameter  of  the  tank  is  GO  inches  and  the  length  is  25  feet.  How 
many  gallons  of  gasoline  are  there  in  the  tank? 

Solution : 

15  divided  by  60  is  0.25 

In  Table  No.  23,  the  area  corresponding  to  0.25  is  0.153546. 
Area  of  cross  section  of  the  gasoline  is  60  X  60  X  0.153546  = 
552.7656  square  inches. 

Twenty-five  feet  is  300  inches  ;  the  tank  contains  300  X 
552.7656  —  165829.68  cubic  inches.  One  gallon  is  231  cubic 
inches.  The  tank  contains  165829.68  divided  by  231  =  717.88, 
or  718  gallons. ' 

NOTE. — If  the  tank  is  more  than  half  full,  figure  first  the 
cubical  contents  of  the  whole  tank  if  full,  then  figure  the  cubical 
contents  of  the  empty  space  and  subtract  the  last  quantity  from 
the  first,  and  the  difference  is  the  cubical  contents  of  the  fluid 
in  the  tank. 

Circular  Lune. 

The  circular  lune  is  a  crescent-shaped 
figure  bounded  by  two  arcs,  as  a  b  c  and 
adc.  (Fig.  8). 

Its  area  is  obtained  by  first  finding 
the  area  of  the  segment  a  dc  (having  c% 
for  center  of  the  circle),  then  the  area 
of  the  segment  a  b  c  (having  c\  for  center 
of  circle),  then  by  subtracting  the  area  of 
the  last  segment  from  the  area  of  the  first ; 
the  difference  is  the  area  of  the  lune. 

A  practical  example  of  a  circular  lune  is  the  area  of  the 
opening  in  a  straight-way  valve  when  it  is  partly  shut. 


FIG.  9. 


Circular  Zone. 


Circular  Zone. 

The  shaded  part,  a  b  c  d,  of  the  figure 
is  called  a  circular  zone.  Its  area  is  ob- 
tained by  first  finding  the  area  of  the 
circle  and  then  subtracting  the  area  of  the 
two  segments  ;  the  difference  is  the  area  of 
the  zone.  When  the  zone  is  narrow  in 
proportion  to  the  diameter,  its  area  is  ob- 
tained very  nearly  by  following  the  rule : 
Add  line  a  b  or  c  d  to  the  diameter  of  the 
circle,  divide  the  sum  by  2  and  multiply 


MENSURATION.  203 

the  quotient  by  the  width  of  the  zone,  and  the  product  is  the 
area. 

To  Compute  the  Volume  of  a  Segment  of  a  Sphere. 

RULE.  FIG.  10. 

Square  half  the  length  of  its  base,  and        |« 8- -I    , 

multiply  by  3.     To  this  product  add  square  ^^— ^—  \-jf- 

of  the  height.     Multiply  the  sum  by  the  "\ 

height  and  by  0.5230.  /\^          /\1 

EXAMPLE.  ( 

Find  volume  of  the  spherical  segment      \ 
shown   in    Fig.  10;    base   line   is  8"   and        \ 
height  is  2". 

Solution  :  Segment  of  a  Sphere. 

Volume  =  v  =  (3  X  42  -f  22)  X  2  X  0.5236 
v  =  (3  X  16  +  4)  X  2  X  0.5236 
v  =  52  X  2  X  0.5236 
v  =  54.4544  cubic  inches. 

To  Find  the  Volume  of  a  Spherical  Segment,  when  the 

Height  of  the  Segment  and  the  Diameter  of 

the  Sphere  are  Known. 

RULE. 

Multiply  the  diameter  of  sphere  by  3,  and  from  this  product 
subtract  twice  the  height  of  segment.  Multiply  the  remainder 
by  the  square  of  the  height  and  the  product  by  0.5236. 

EXAMPLE. 

The  segment  (Fig.  10)  is  cut  from  a  sphere  10  inches  in 
diameter  and  it  is  2  inches  high.  Figure  it  by  this  last  rule. 

Solution : 

Volume  =  v  =  (10  X  3  —  2  X  2)  X  22  X  0.5236 
v  =  (30  —  4)  X  4  X  0.5236 
v  =  26  X  4  X  0.5236 
i/  =  54.4544  square  inches. 

To  Find  the  Surface  of  a  Cylinder. 

RULE. 

Multiply  the  circumference  by  the  length,  and  to  this  pro- 
duct add  the  area  of  the  two  ends. 

A  cylinder  has  the  largest  volume  with  the  smallest  surface 
when  length  and  diameter  are  equal  to  each  other. 


2O4 


MENSURATION. 


To  Find  the  Volume  of  a  Cylinder. 

RULE. 

Multiply  area  of  end  by  length  of  cylinder,  and  the  product 
is  the  volume  of  the  cylinder. 
EXAMPLE. 

What  is  the  volume  of  a  cylinder  4  inches  in  diameter  and  9 
inches  long? 

Solution : 

Area  of  end  —  r2  TT 
Volume  =  r2  TT  /—  2  X  2  X  3.1416  X  9  =  113.0976  cubic  inches. 

To  Find  the  Solid  Contents  of  a  Hollow  Cylinder. 

RULE. 

Find  area  of  end  according  to  outside  diameter ;  also  find 
area  according  to  inside  diameter  ;  subtract  the  last  area  from 
the  first  and  multiply  the  difference  by  the  length  of  the 
cylinder. 

Formula : 

Area  =  (/?2  —  r2)  TT  / 
R  =  Outside  radius. 
r  —  Inside  radius. 
/  =  Length  of  cylinder. 
EXAMPLE. 

Find  the  solid  contents  of  a  hollow  cylinder  of  6  feet  outside 
diameter,  4  feet  inside  diameter  and  5  feet  long. 
Solution : 

Solid  contents  =  x  —  (32  —  22)  X  3.1416  X  5 
x  —  (9  —  4)  X  3.1416  X  5 
x-  —  5  X  3.1416  X  5 
x  =  78.54  cubic  feet. 

FIG.  10. 

^--~ 

To  Find  the  Area  of  the  Curved 
Surface  of  a  Cone. 

(See  Fig.  10). 

RULE. 

Multiply  the  circumference 
of  the  base  by  the  slant  height 
and  divide  the  product  by  2  ;  the 
quotient  is  the  area  of  the  curved 
surface.  If  the  total  surface  is 
wanted,  the  area  of  the  base  is 
added  to  the  curved  area. 


Cone, 


MENSURATION. 


205 


If  the  perpendicular  height  is  known,  the  length  of  the  slant 
side  or  the  slant  height  is  found  by  adding  the  square  of  the  per- 
pendicular height  to  the  square  of  the  radius  and  extracting  the 
square  root  of  the  sum. 

Formula : 


Curved  area  =  x  = 


r  =•  Radius  of  base. 
*/=  Diameter  of  base. 
h  =  Perpendicular  height. 

To  Find  the  Volume  of  a  Cone. 

RULE. 

Multiply  the  area  of  the  base  by  the  perpendicular  height, 
and  divide  the  product  by  3. 

By  formula : 

Volume  =    r*"h 


To  Find  the  Area  of  the  Curved  Surface  of  a  Frustum 
of  a  Cone. 

(See  Fig.  11). 

RULE. 

Add  circumference  of  small  end  to 
circumference  of  large  end,  multi- 
ply this  sum  by  the  slant  height  and 
divide  the  product  by  2. 

Formula : 

Curved  area  =  (21?ir  +  2r  *)    — 
which  reduces  to 

Curved  area  =  (^?  -f-  r)  *  S 

If  the  perpendicular  height  instead  of 
the  slant  height  is  known,  we  have : 

Curved  area  =  (R  -}-  r)  K 

R  =  Large  radius. 
r  =  Small  radius. 


Frustum  of  a  Cone. 


_  r^  _j_  j 


h  =  Perpendicular  height, 
^  =  Slant  height, 


206 


MENSURATION. 


To  Find  the  Volume  of  a  Frustum  of  a  Cone. 

RULE. 

Square  the  largest  radius ;  square  the  smallest  radius. 
Multiply  largest  radius  by  smallest  radius  ;  add  these  three  pro- 
ducts and  multiply  their  sum  by  3.1416 ;  multiply  this  last 
product  by  one-third  of  the  perpendicular  height. 

Formula : 

Volume  =  (R*  -f  r2  -f-  Rr)  K  _A_ 

3 

EXAMPLE. 

Find  the  volume  of  a  frustum  of  a  cone.  The  largest 
diameter  is  6  feet,  the  smallest  diameter  is  4  feet,  and  perpen- 
dicular height  is  12  feet. 

Solution: 

Volume  =  x  —  (32  +  22  +  3  X  2)  X  3.1416  X  — 

3 

x  —  (9  -f  4  -f-  6)  X  3.1416  X  4 
x  =  19  X  3.1416  X  4 
x  —  238.7616  cubic  feet. 

NOTE. — This  rule  will  also  apply  for  finding  the  solid  con- 
tents of  wood  in  a  log. 


12. 


Pyramid 


To  Find  the  Area  of  the  Slanted 
Surface  of  a  Pyramid. 

(See  Fig.  12). 

RULE. 

Multiply  the  length  of  the  perim- 
eter of  the  base  by  the  slant  height  of 
the  side  (not  the  slant  height  of  the 
edge).  Divide  the  product  by  2,  and 
the  quotient  is  the  area. 


To  Find  the  Total  Area  of  the  Surface  of  a  Pyramid. 

RULE. 

Find  area  of  the  slanted  surface  as  explained  above,  and  to 
this  add  the  area  of  a  polygon  equal  to  the  base  of  the  pyramid. 


To  Find  the  Volume  of  a  Pyramid. 


RULE. 


Multiply  the  area^  of  the  base  by  one-third  of  the  perpen- 
dicular height 


MENSURATION. 


207 


frlG.  13, 


Frustum  Of  a  Pyramid 


To  Find  the  Area  of  the  Slanted 

Surface  of  a  Frustum  of 

a  Pyramid. 

RULE. 

Add  perimeter  of  the  small  end 
to  the  perimeter  of  the  large  end. 
Multiply  this  sum  by  the  slant  height 
of  the  side  (not  slant  height  of  edge). 
Divide  the  product  by  2. 


To  Find  the  Total  Area  of  the  Surface  of  a  Frustum  of  a 
Pyramid. 

RULE. 

Find  the  area  of  the  slanted  surface  as  explained  above, 
and  to  this  area  add  the  area  of  the  two  ends.  Their  areas  are 
obtained  in  the  same  way  as  areas  of  polygons.  (See  page  196). 

To  Compute  the  Volume  of  a  Frustum  of  a  Pyramid. 

RULE. 

Multiply  the  area  of  the  small  end  by  the  area  of  the  large 
end,  extract  the  square  root  of  the  product,  and  to  this  add  the 
area  of  the  small  end  and  the  area  of  the  large  end  ;  multiply  the 
sum  by  one-third  of  the  perpendicular  height. 

Formula  : 

Volume  =  A  (a  +  A  -f  \f~A~a) 

3 

EXAMPLE. 

Find  volume  of  a  frustum  of  a  pyramid  when  the  area  of 
the  small  end  is  8  square  feet,  the  area  of  the  large  end  is  18 
square  feet  and  the  perpendicular  height  is  30  feet. 


Of\ 

Volume  =  v=  - 


18_j_  x/18  X  8  ) 


v  =  WX  (  8  +  18  +       l44~  ) 

v  =  10  X  (  8  -f-  18  -f  12  ) 

•v  =  10  X  38 

v  =  380  cubic  feet. 

To  Find  the  Surface  of  a  Sphere. 

RULE. 

Multiply  the  circumference  by  the  diameter. 

NOTE.  —  The  surface  of  a  sphere  is  equal  to  the  curved  sur- 
face of  a  cylinder  having  diameter  and  length  equal  to  the 
diameter  of  the  sphere. 


208  MENSURATION. 

To  Find  the  Volume  of  a  Sphere. 

RULE. 

Multiply  the  cube  of  the  diameter  by  3.1416,  divide  the 
product  by  6  and  the  quotient  is  the  volume  of  the  sphere.  Or, 
another  rule  is :  Multiply  the  cube  of  the  diameter  by  0.5236  and 
the  product  is  the  volume  of  the  sphere. 

EXAMPLE. 

Find  the  volume  of  a  sphere  15''  diameter. 

Solution : 

0.5236  X  15  X  15  X  15  =  1767.15  cubic  inches. 

A  sphere  twice  as  large  in  diameter  as  another  has  twice  the 
circumference,  four  times  the  surface,  eight  times  the  volume, 
and  if  of  the  same  material  will  weigh  eight  times  as  much. 

To  Compute  the  Diameter  of  a  Sphere  when  the  Volume 
is  Known. 

RULE. 

Divide  the  volume  by  0.5236  and  the  cube  root  of  the 
quotient  is  the  diameter  of  the  sphere. 

To  Compute  the  Circumference  of  an  Ellipse. 

RULE. 

Add  the  square  of  the  largest  diameter  to  the  square  of  the 
smallest  diameter  and  divide  the  sum  by  2;  multiply  the  square 
root  of  the  quotient  by  3.1416.  * 

EXAMPLE. 

Find  the  circumference  of  an  ellipse.  The  largest  diameter 
is  24  inches  and  the  smallest  diameter  is  18  mches. 

Solution : 
Circumference  =  c=  3.1416 

c  =  3.1416^450 
c—  3.1416  X  21.2132 
c  —  66.643  inches. 

To  Compute  the  Area  of  an  Ellipse. 

RULE. 

Multiply  the  smallest  diameter  by  the  largest  diameter,  and 
this  product  by  0.7854. 

*  This  Rule  gives  only  approximate  results.    There  is  no  known  rule  giving 
exact  results. 


CIRCUMFERENCES    AND   AREAS    OF    CIRCLES. 


209 


TABLE  No.  24.  —Giving  Circumferences  and  Areas  of 
Circles. 


Diameter. 

Circumfer- 

Area. 

Diameter. 

Circumfer- 

Area. 

ence. 

ence. 

, 

0.0491 

0.00019 

H 

2.1108 

0.35454 

l 

0.0982 

0.00077 

iff 

2.1598 

0.37122 

A 

0.1473 

0.00173 

6¥ 

2.2089 

0.38829 

A 

0.1904 

0.00307 

|| 

2.2580 

0.40574 

A 

0.2454 

0.00479 

|| 

2.3071 

0.42357 

A 

0.2945 

0.00690 

X 

2.3562 

0.44179 

i  ^ 

0.3430 

0.00940 

ft 

2.4053 

0.40039 

0.3927 

0.01227 

if 

2.4544 

0.47937 

'^ 

0.4418 

0.01553 

1? 

2.5035 

0.49874 

A 

0.4909 

0.01918 

it 

2.5525 

0.51849 

H 

0.5400 

0.02320 

If 

2.6016 

0.53862 

A 

0.5890 

0.02761 

H 

2.6507 

0.55914 

if 

0.0381 

0.03241 

If 

2.6998 

0.58004 

A 

0.0872 

0.03758 

2.7489 

0.60132 

I! 

0.7303 

0.04314 

'ft 

2.7980 

0.62299 

X 

0.7854 

0.04909 

2.8471 

0.64504 

H 

0.8345 

0.05542 

tf 

2.8962 

0.66747 

0.8830 

0.06213 

it 

2.9452 

0.69029 

if 

0.9327 

0.00922 

H 

2.9943 

0.71349 

fV        !    0.9818 

0.07070 

3.0434 

0.73708 

j?lr        1.0308 

0.0845(5 

I! 

3.0925 

0.76105 

U        1.0799 

0.09281 

3.1410 

0.78540 

]    1.1290 

0.10144 

IST 

3.1907 

0.81013 

H            1-1781 

0.11045 

1  j 

3.2398 

0.83525 

15-        1.2272 

0.11984 

l^3f 

3.2889 

0.86075 

J3         1.27(53 

0.12902 

1  r<> 

3.3379 

0.88664 

1.8254 

0.13979 

1^¥ 

3.3870 

0.91291 

rt            1.3744 

0.15033 

3.4361 

0.93956 

S£        1.4235 

0.10120 

IA 

3.4852 

0.96660 

11        1.4720 

0.17258 

1^6 

3.5343 

0.99402 

;!1        1.5217 

0.18427 

1    9 

3.5834 

1.02182 

>^            1.5708 

0.19035 

IA 

3.0325 

1.05001 

-21-        1.0199 

0.20881 

18 

3.6816 

1.07858 

11        1.0090 

0.22100 

iA 

3.7306 

1.10753 

||        1.7181 

0.23489 

i*l 

3.7797 

1.13687 

A        i    1-7671 

0.24850 

3.8288 

1.16659 

H        1.8162 

0.20250 

i^f 

3.8779 

1.19670 

if 

1.8653 

0.27088 

IX 

3.9270 

1.22718 

H 

1.9144 

0.29165 

1^1 

3.9761 

1.25806 

* 

1.9635 

0.30680 

1A 

4.0252 

1.28931 

ft 

2.0126 

0.32233 

iff 

4.0743 

1.32095 

H    i    2.0017 

0.33824 

4.1233 

1.35297 

210          CIRCUMFERENCES    AND    AREAS    OF    CIRCLES. 


Diameter. 

Circumfer- 
ence. 

Area. 

Diameter. 

Circumfer- 
ence. 

Area. 

m 

4.1724 

1.38538 

2^ 

6.6759 

3.5466 

4.2215 

1.41817 

2fV 

6.8722 

3.7584 

ill 

4.2706 

1.45134 

2X 

7.0686 

3.9761 

iH 

4.3197 

1.48489 

2T56 

7.2649 

4.2 

ill 

4.3688 

1.51883 

2^ 

7.4613 

4.4301 

4.4179 

1.55316 

2r75 

7.6576 

4.6664 

li 

4.4670 

1.58786 

2^ 

7.8540 

4.9087 

T7ff 

4.5160 

1.62295 

2T96 

8.0503 

5.1573 

•    19 

4.5651 

1.65843 

2ft 

8.2467 

5.4119 

-  M 

4.6142 

1.69428 

2H 

8.4430 

5.6727 

it 

4.6633 

1.73052 

2% 

8.6394 

5.9396 

|| 

4.7124 

1.76715 

2|| 

8.8357 

6.2126 

li 

4.7615 

1.80415 

2# 

9.0321 

6.4918 

1  7 

4.8106 

1.84154 

2J-5 

9.2284 

6.7772 

if 

4.8597 

1.87932 

3 

9.4248 

7.0686 

T9* 

4.9087 

1.91748 

8& 

9.6211 

7.3662 

if 

4.9578 

1.95602 

3^ 

9.8175 

7.6699 

it 

3  2 

5.0069 

1.99494 

B& 

10.0138 

7.9798 

II 

5.0560 

2.03425 

3X 

10.2102 

8.2958 

# 

5.1051 

2.07394 

3,5,; 

10.4066 

8.6179 

111 

5.1542 

2.11402 

3^ 

10.6029 

8.9462 

If* 

5.2033 

2.15448 

3TV 

10.7992 

9.2807 

Mi 

5.2524 

2.19532 

3^ 

10.9956 

9.6211 

ill 

5.3014 

2.23654 

3r9. 

11.1919 

9.9678 

ill 

5.3505 

2.27815 

3^ 

11.3883 

10.3206 

ill 

5.3996 

2.32015 

8H 

11.5846 

10.6796 

iff 

5.4487 

2.36252 

3K 

11.7810 

11.0447 

iK 

5.4978 

2.40528 

3H 

11.9773 

11.4160 

if! 

5.5469 

2.44843 

3% 

12.1737 

11.7933 

iff 

5.5960 

2.49195 

3}| 

12.3701 

12.1768 

iff 

5.6450 

2.53586 

4 

12.5664 

12.5664 

ifl 

5.6941 

2.58016 

4TV 

12.7628 

12.9622 

iff 

5.7432 

2.62483 

4^ 

12.9591 

13.3641 

ill 

5.7923 

2.66989 

4T3«3 

13.1554 

13.7721 

ill 

5.8414 

2.71534 

4^ 

13.3518 

14.1863 

1$ 

5.8905 

2.76117 

4r5,T 

13.5481 

14.6066 

ill 

5.9396 

2.80738 

4^ 

13.7445 

15.0330 

i?! 

5.9887 

2.85397 

4rV 

13.9408 

15.4656 

iff 

6.0377 

2.90095 

4^ 

14.1372 

15.9043 

if! 

6.0868 

2.94831 

4r9,5 

14.3335 

16.3492 

Jff 

6.1359 

2.99606 

4>/8 

14.5299 

16.8002 

HI 

6.1850 

3.04418 

4H 

14.7262 

17.2573 

iff 

6.2341 

3.0927 

4^: 

14.9226 

17.7206 

2 

6.2832 

3.1416 

4}| 

15.1189 

18.19 

2A 

6.4795 

3.3410 

4% 

15.3153 

18.6655 

CIRCUMFERENCES    AND    AREAS    OF    CIRCLES. 


211 


Diameter 

Circumfer- 
ence. 

Area. 

Diameter 

Circumfer- 
ence. 

Area. 

HI 

15.5116 

19.1472 

10# 

32.9868 

86.5903 

5 

15.7080 

19.6350 

10# 

33.3795 

88.6643 

5M 

16.1007 

20.6290 

IOK 

33.7722 

90.7625 

5% 

16.4934 

21.6476 

10^ 

34.1649 

92.8858 

5# 

16.8861 

22.6907 

11 

34.5576 

95.0334 

5^ 

17.2788 

23.7583 

HH 

34.9503 

97.2055 

5# 

17.6715 

24.8505 

nx 

35.343 

99.4019 

5% 

18.0642 

25.9673 

ii# 

35.7357 

101.6234 

5% 

18.4569 

27.1084 

HM 

36.1284 

103.8691 

6 

18.8496 

28.2744 

IIH 

36.5211 

106.1394 

6^ 

19.2423 

29.4648 

ii# 

36.9138 

108.4338 

6% 

19.635 

30.6797 

11% 

37.3065 

110.7537 

6H 

20.0277 

31.9191 

12 

37.6992 

113.098 

6^ 

20.4204 

33.1831 

12tf 

38.4846 

117.859 

6^ 

20.8131 

34.4717 

12/2 

39.2700 

122.719 

6% 

21.2058 

35.7848 

12* 

40.0554 

127.677 

6% 

21.5985 

37.1224 

13 

40.8408 

132.733 

21.9912 

38.4846 

18# 

41.6262 

137.887 

^ 

22.3839 

39.8713 

13>^ 

42.4116 

143.139 

7% 

22.7766          41.2826  - 

18M 

43.1970 

148.490 

7>i 

23.1693 

42.7184 

14 

43.9824 

153.938 

7^ 

23.5620 

44.1787 

14X 

44.7678 

159.485 

7>g 

23.9547          45.6636 

U/2 

45.5532 

165.130 

7% 

24.3474          47.1731 

14% 

46.3386 

170.874 

7/8 

24.7401          48.7071 

15 

47.1240 

176.715 

8 

25.132S          50.2656 

15X 

47  9094 

182.655 

8^ 

25.52:.r>          51.8487 

15^ 

48.6948 

188.692 

8% 

2.->.  1)1  82          53.4561 

15* 

49.4802 

194.828 

8^ 

26.3109  ;       55.0884 

16 

50.2656 

201.062 

S/2 

26.7086 

56.7451 

16^ 

51.051 

207.395 

8^ 

27.0963 

58.4264 

16^ 

51.8364 

213.825 

8% 

27.489 

60.1319 

16% 

52.6218 

220.354 

8/8 

27.8817          61.8625 

17 

53.4072 

226.981 

9 

28.2744 

63.6174 

17X 

54.1926 

233.706 

$% 

28.6671 

65.3968 

17^ 

54.9780 

240.529 

9% 

29.0598 

(57.2008 

17% 

55.7634 

247.450 

9/8 

29.  4525 

69.0293 

18 

56.5488 

254.470 

9K 

29.8452 

70.8823 

18% 

57.3342 

261.587 

9^ 

30.2379 

72.7599 

18# 

58.1196 

268.803 

9% 

30.6306 

74.6619 

18% 

58.905 

276.117 

9% 

31.0233 

76.5888 

19 

59.6904 

283.529 

10 

31.4160 

78.5400 

19% 

60.4758 

291.040 

10# 

31.8087 

80.5158 

19^ 

61.2612 

298.648 

iox 

32.2014 

82.5158 

19% 

62.0466 

306.355 

10^ 

32.5941 

84.5409 

20 

62.8320 

314.16 

212  CIRCUMFERENCES  ANt>   AREAS   OF   CIRCLES. 


Diameter 

Circumfer- 
ence. 

Area. 

Diameter 

Circumfer- 
ence. 

Area. 

21 

65.9736 

346.361 

66 

207.34 

3421.19 

22 

69.1152 

380.134 

67 

210.49 

3525.65 

23 

72.2568 

415.477 

68 

213.63 

3631.68 

24 

75.3984 

452.39 

69 

216.77 

3739.28 

25 

78.540 

490.87 

70 

219.91 

3848.45 

26 

81.681 

530.93 

71 

223.05 

3959.19 

27 

84.823 

572.56 

72 

226.19 

4071.50 

28 

87.965 

615.75 

73 

229.34 

4185.39 

29 

91.106 

660.52 

74 

232.48 

4300.84 

30 

94.248 

706.86 

75 

235.62 

4417.86 

31 

97.389 

754.77 

76 

238.76 

4536.46 

32 

100.53 

804.25 

77 

241.90 

4656.63 

33 

103.67 

855.30 

78 

245.04 

4778.36 

34 

106.81 

907.92 

79 

248.19 

4901.67 

35 

109.96 

962.11 

80 

251.33 

5026.55 

36 

113.10 

1017.88 

81 

254.47 

5153.00 

37 

116.24 

1075.21 

82 

257.61 

5281.02 

38 

119.38 

1134.11 

83 

260.75 

5410.61 

39 

122.52 

1194.59 

84 

263.89 

5541.77 

40 

125.66 

1256.64 

85 

267.04 

5674.50 

41 

128.81 

1320.25 

86 

270.18 

5808.80 

42 

131.95 

1385.44 

87 

273.32 

5944.08 

43 

135.09 

1452.20 

88 

276.46 

6082.12 

44 

138.23 

1520.53 

89 

279.60 

6221.14 

45 

141.37 

1590.43 

90 

282.74 

6361.73 

46 

144.51 

1661.90 

91 

285.88 

6503.88 

47 

147.65 

1734.94 

92 

289.03 

G647.61 

48 

150.80 

1809.56 

93 

292.17 

6792.91 

49 

153.94 

1885.74 

94 

295.31 

6939.78 

50 

157.08 

1963.50 

95 

298.45 

7088.22 

51 

160.22 

2042.82 

96 

301.59 

7238.23 

52 

163.36 

2123.72 

97 

304.73 

7389.81 

53 

166.50 

2206.18 

98 

307.88 

7542.96 

54 

169.65 

2290.22 

99 

311.02 

7697.69 

55 

172.79 

2375.83 

100 

314.16 

7853.98 

56 

175.93 

2463.01 

101 

317.30    8011.85 

57 

179.07 

2551.76 

102 

320.44    8171.28 

58 

182.21 

2642.08 

103 

323.58  i   8332.29 

59 

185.35 

2733.97 

104 

326.73  1   8494.87 

60 

188.50 

2827.43 

105 

329.87 

8659.01 

61 

191.64 

2922.47 

106 

333.01 

8824.73 

62 

194.78 

3019.07 

107 

336.15 

8992.02 

63 

197.92 

3117.25 

108 

339.29 

9160.88 

64 

201.06 

3216.99 

109 

342.43 

9331.32 

65 

204.20 

3318.31 

110 

345.58 

9503.32 

Strength  of  flfoaterfals. 


The  strength  of  materials  may  be  divided  into  Tensile, 
Crushing,  Transverse,  Torsional,  or  Shearing,  and  besides 
this,  the  elasticity  of  the  material  or  its  resistance  against 
deflection  must  also  be  taken  into  consideration  in  figuring  for 
strength. 

Tensile  Strength. 

From  experiments  it  is  known  that  it  it  will  take  from 
40,000  to  70,000  pounds  to  tear  off  a  bar  of  wrought  iron  one 
inch  square.  Therefore  we  usually  say  that  the  tensile  strength 
of  wrought  iron  is  from  40,000  to  70,000  pounds,  according  to 
quality.  The  average  is  50,000  to  55,000  pounds.  The  tensile 
strength  of  any  body  is  in  proportion  to  its  cross  sectional  area ; 
thus,  if  a  bar  of  iron  of  one  square  inch  area  will  pull  asunder 
under  a  load  of  40,000  pounds,  it  will  take  80,000  pounds  to  pull 
asunder  another  bar  of  the  same  kind  of  iron  but  of  two  square 
inches  area.  The  tensile  strength  is  independent  of  the  length 
of  the  bar,  if  it  is  not  so  long  that  its  own  weight  must  be 
taken  into  consideration.  Table  No.  24  gives  the  load  which 
will  pull  asunder  one  square  inch  of  the  most  common  materials. 

No  part  of  any  machine  should  be  strained  to  that  limit.  A 
high  factor  of  safety  must  be  used,  sometimes  from  4  to  30  or 
even  more,  which  will  depend  upon  the  kind  of  stress  the  mem- 
ber is  exposed  to,  as  dead  load,  variable  load,  shocks,  etc.  Dif- 
ferent factors  of  safety  are  also  used  for  different  kinds  of 
material.  (See  page  274). 

flodulus  of  Elasticity. 

The  modulus  of  elasticity  for  any  kind  of  material  is  usually 
defined  as  the  amount  of  force  which  would  be  required  to 
stretch  a  straight  bar  of  one  square  inch  area  to  double  its 
length  or  compress  it  to  nothing,  if  this  were  possible.  But  a 
more  comprehensive  definition  is  to  say  that  the  modulus  of 
elasticity  is  the  reciprocal  of  the  fractional  part  of  the  length 
which  one  unit  of  force  will,  within  elastic  limit,  stretch  or  com- 
press one  unit  of  area.  For  instance,  if  the  modulus  of  elasticity 
for  a  certain  kind  of  wrought  iron  is  25,000,000,  it  means  that  it 
would  take  25,000,000  pounds  of  pulling  force  to  stretch  a  bar  of 

(213) 


214  STRENGTH  OF  MATERIALS. 

one  square  inch  area  to  double  its  length,  if  this  could  possibly 
be  done ;  but  it  means  also — which  is  exactly  equivalent — 
that  one  pound  of  pulling  force  will  stretch  a  bar  of  one  square 
inch  area  one  25-millionth  part  of  its  length,  or  one  pound 
compressive  force  will  shorten  the  same  bar  one  25-millionth 
part  of  its  length,  and  that  two  pounds  of  force  will  stretch  or 
compress  twice  as  much,  three  pounds  thrice  as  much,  etc. 

Strength  of  Wrought  Iron. 

From  experiments  it  is  known  that  wrought  iron  can  not 
very  well  be  stretched  or  compressed  more  than  one-thousandth 
part  of  its  length  without  destroying  its  elasticity ;  therefore  if 
a  bar  of  wrought  iron  has  25,000,000  as  its  modulus  of  elas- 
ticity, one  pound  will  stretch  it  ^5 o oV OOIF  of  its  length  and  it 
would  take  25,000  pounds  to  stretch  it  ysVo  of  itslength.  Thus, 
25,000  pounds  would  then  be  said  to  be  its  strength  at  the  limit 
of  elasticity  for  that  kind  of  iron ;  80  to  100  per  cent,  more  will 
usually  be  the  ultimate  breaking  load. 

The  pull  or  load  which  such  a  bar  can  sustain  with  safety 
will  depend  a  great  deal  on  circumstances,  but  it  must  never 
exceed  25,000  pounds  per  square  inch  of  area.  It  must  not 
even  approach  this  limit  if  the  structure  is  of  any  importance  or 
if  the  load  is  to  be  sustained  for  any  lengtty  of  time,  or  if  it  is, 
besides  the  load,  also  exposed  to  shocks  or  jar. 

Strength  of  Cast  Iron. 

Cast  iron  of  good  quality  has  a  modulus  of  elasticity  of 
15,000,000  pounds,  but  if  strained  so  it  will  stretch  y^o  o  of  its 
length  its  elasticity  is  usually  destroyed.  For  instance,  a  bar  of 
cast  iron  of  one  square  incn  area  is  exposed  to  tensile  strain, 
its  modulus  of  elasticity  being  15,000,000  pounds  and  its  elas- 
ticity being  destroyed  if  it  stretches  ysV o  of  its  length,  what 
then  would  be  its  strength  at  limit  of  elasticity?  One  pound 
will  stretch  it  ^-.^^^  of  its  length,  therefore  it  must  take 
10,000  pounds  to  stretch  it  ysW  of  its  length;  thus  we  would 
say  that  10,000  pounds  is  its  strength  at  limit  of  elasticity.  It 
is  not  always  that  cast  iron  is  of  as  good  quality  as  that;  very 
frequently  its  elasticity  is  destroyed  if  it  is  exposed  to  a  tensile 
stress  of  6,000  pounds  per  square  inch  of  area;  thus  the  strength 
of  cast  iron  at  its  limit  of  elasticity  is  often  found  to  be  only 
6,000  pounds  instead  of  10,000  pounds.  Besides,  it  is  very 
often  found  that  a  pulling  force  of  10,000  pounds  will  stretcn 
a  bar  of  one  square  inch  area  one  twelve-hundredth  part  of  its 
length,  and  this,  of  course,  gives  the  modulus  of  elasticity 
12,000,000  pounds.  Frequently  cast  iron  is  of  such  quality  that 
it  cannot  be  stretched  over  ^^^  of  its  length  before  its  elas- 
ticity is  destroyed.  Cast  iron  is  very  variable  in  quality,  and 


STRENGTH    OF    MATERIALS.  215 

especially   so   with   regard   to  its  tensile  strength.     Generally 
speaking,  we  may  say  that  for  cast  iron  the 

Modulus  of  elasticity  is  12,000.000  to  15,000,000  pounds. 

Tensile  strength  at  limit  of  elasticity,  5,000  to  10,000  pounds. 

Ultimate  tensile  strength,  10,000  to  20,000  pounds. 

Elongation  Under  Tension. 

The  total  stretch  or  elongation  of  any  specimen  when 
exposed  to  tensile  stress  within  the  elastic  limit  is  directly  pro- 
portional to  the  length  of  the  specimen,  but  it  is  inversely 
proportional  to  the  modulus  of  elasticity  and  the  cross  sectional 
area  of  the  specimen.  The  following  formulas  may,  therefore, 
be  used  in  such  calculations  : 

F  _  P  X   L  p  _  E  X  s  X  A 

s'xTA  L 

P  X   L  A  —  PX  L 

£~y^4~  ~7x~E 

_  E  X  sX  A 


E  =  Modulus  of  elasticity  in  pounds  per  square  inch. 
P  —  Load  or  force  in  pounds  acting  to  elongate  the  specimen. 
s  —  Total  stretch  of  specimen  in  inches  in  the  length  /,. 
L  —  Original  length  of  specimen  in  inches  before  force  is 

applied. 

A  —  Cross-sectional  area  of  specimen  in  square  inches. 
EXAMPLE. 

From  experiments  it  is  known  that  the  modulus  of  elasticity 
for  a  certain  kind  of  wrought  iron  is  28,000,000;  what  will  then 
be  the  total  stretch  or  elongation  in  a  round  boiler  stay,  \% 
inches  in  diameter  and  6  feet  long,  when  exposed  to  a  stress  of 
5000  pounds? 
Solution  : 

1  14  inches  diameter  =  1.227  inches  area  (see  table,  page  209) 
C  feet  long  =•  72  inches. 
P  X  L 


s  = 


X  A 
5000  X  72 


28000000  X  1227 

s  =  0.0105  inches  =  total  stretch  in  the  stay. 

NOTE.  —  As  already  stated,  wrought  iron  can  not  be 
stretched  as  much  as  one-thousandth  part  of  its  original  length 
without  danger  of  destroying  its  elasticity;  thus,  for  this  stay, 
which  is  72  inches,  the  limit  of  elasticity  will  be  at  a  stretch  of 
0.072  inches ;  therefore  the  stretch  produced  by  a  load  of  5,000 
pounds,  which  is  calculated  to  be  0.0105  inches,  is  well  within 
the  safe  limit. 


2l6 


STRENGTH    OF   MATERIALS. 


TABLE    No.    25.— Modulus    of    Elasticity    and    Ultimate 
Tensile  Strength  of  Various  Materials. 


MATERIALS. 

Modulus  of 
Elasticity  in 
Pounds  per 
Square  Inch. 

Ultimate 
Tensile 
Strength  in 
Pounds  per 
Square  Inch. 

Modulus  of 
Elasticity  in 
Kilograms  per 
Square 
Centimeter. 

Ultimate 
Tensile 
Strength  in 
Kilograms 
per  Square 
Centimeter. 

Cast  steel    .       .    . 
Bessemer  steel  .    . 
Wrought  iron  bars 
Wrought  iron  wire 

Cast  iron  *.'..< 

Copper  bolts  .    .    . 
Brass   

30,000,000 
28.000,000 
25,000,000 
28,000,000 
12,000,000 
to 
15,000,000 
18,000,000 
9,000,000 

100,000 
70,000 
55,000 
75,000 
10,000 
to 
20,000 
35,000 
17,700 

2,200,000 
1,970,000 
1,700,000 
1,970,000 
80,0000 
to 
1,000,000 
1,200,000 
630,000 

7,000 
4,930 
3,850 
5,250 
700 
to 
1,400 
2,400 
1,200 

Oak  

,500,000 

17,000 

105,000 

1,200 

Hickory      .... 
Maple  

,400,000 
,100,000 

20,000 
15,000 

98,000 
77,000 

1,400 
1,000 

Pitch  pine  .... 
Pine     

,600,000 
,100,000 

15,000 
10,000 

112,000 

77,000 

1,000 
700 

Soruce 

,100,000 

10,000 

77,000 

700 

The  two  last  columns  in  above  table  are  calculated  by  the 
rule  :  One  pound  per  sq.  inch  —  0.07031  kilograms  pei  sq.  centi- 
meter and  the  result  is  reduced  to  the  nearest  round  number. 


Formulas  for  Tensile  Strength. 

The  ultimate  tensile  strength  of  any  specimen  is  in  propor- 
tion to  its  cross-sectional  area,  and  is  expressed  by  the  following 
formula : 

P  =  A  X  S 


Side  of  a  square  bar 


-v? 


^  _        P  Diameter  of  a  round  bar  —  'V— ; ^ 

o 
P  =  Force  in  pounds  which  will  pull  the  specimen  asunder. 

S  =  Ultimate  tensile  strength  in  pounds  per  square  inch. 
(See  Table  No.  25). 

A  —  Cross-sectional  area  of  the  specimen  in  square  inches. 


*  Very  strong  cast  iron  may  have  an  ultimate  tensile  strength  as  high  as  30,000 
pounds  per  square  inch. 


STRENGTH  OF  MATERIALS.  21  7 

EXAMPLE. 

A  piece  of  iron  l/2  inch  square  is  tested  in  a  testing  machine 
and  breaks  at  a  total  stress  of  14,210  pounds.  What  is  the 
ultimate  tensile  strength  per  square  inch  ? 

Solution  : 

A  bar  y2  inch  square  has  a  cross-sectional  area  of  %"  X  %n 
is  X  square  inch. 

S=  !i?L0  =  56,840  pounds  per  square  inch. 

EXAMPLE. 

What  will  be  the  breaking  load  for  a  wrought  iron  bar 
}i"  X  y&"  when  exposed  to  tensile  stress,  the  ultimate  tensile 
strength  of  the  iron  being  55,000  pounds  per  square  inch,  as 
given  in  Table  No.  25,  page  210  ? 

Solution  : 

A  bar  y%"  X  y%"  is  B9r  square  inches  in  area. 

P  —  fa  x  55,000  =  7734  pounds,  which  will  break  the  bar. 

In  order  to  obtain  the  safe  working  stress  introduce  a  suit- 
able factor  of  safety,  from  5  to  10,  according  to  circumstances, 
and  calculate  by  the  following  formulas  : 

P  Xf=A  X  S  \~P~X~f 

.  v   c  Side  of  a  square  bar  —  \  —  _  _  *. 

P  =  •& 

A  _  P  X  f  Diameter  of  a  round  bar  =  -y         f 

s  ^0.7854  6* 

P  —  Load  in  pounds. 

/•=.  Factor  of  safety. 

EXAMPLE. 

A  load  of  24,000  pounds  is  suspended  on  a  round  wrought 
iron  bar.  The  ultimate  tensile  strength  of  the  iron  is  55,000 
pounds  per  square  inch.  What  should  be  the  diameter  of  the 
bar  to  sustain  the  load,  with  10  as  the  factor  of  safety  ? 

Solution: 


A  =  24000X1°  =  4.363  square  inches. 
55000 

In  Table  No.  24,  we  find  the  nearest  larger  diameter  to  be 
2}i  inches. 

The  diameter  may  also  be  calculated  directly  by  the  fol- 
lowing formula  : 


2l8  STRENGTH   OF   MATERIALS. 


-J/'X/ 

5"  X  0.7854 


^24000  X  10 
55000  X  0.7854 


D  =  2.358,  or  nearly  2^  inches  diameter. 

To  Find  the  Diameter  of  a  Bolt  to  Resist  a  Given  Load. 

RULE. 

Multiply  pull  in  pounds  by  the  factor  of  safety.  Multiply 
the  ultimate  tensile  strength  of  the  material  by  0.7854 ;  divide 
this  first  product  by  the  last  and  extract  the  square  root  from 
the  quotient  which  will  then  be  diameter  of  bolt  at  the  bottom 
of  the  thread. 


P  = 


T  X  0.7854 
X  S  X  0.7854 

D'2  X  S  X  0.7854 


D  =  Diameter  of  bolt  or  screw  in  the  bottom  of  the  thread. 

P  •=•  Load  or  pull  in  pounds. 

y=  Factor  of  safety. 

S  =  Ultimate  tensile  strength  per  square  inch. 


NOTE.  —  Bolts  are  frequently  exposed  to  a  considerable 
amount  of  initial  stress,  due  to  the  tightening  of  nuts,  which 
must  always  be  allowed  for  when  deciding  upon  the  load  to  be 
considered  when  calculating  their  diameter. 

EXAMPLE. 

Find  diameter  of  a  bolt  to  sustain  a  load  of  4,450  pounds, 
taking  10  as  factor  of  safety  and  ultimate  tensile  strength  of  the 
iron  to  be  50,000  pounds  per  square  inch. 

Solution  : 

"4450  X 


_     /"4450 
"  ^ 


50000  X  0.7854 


D  —  1.064"  in  the  bottom  of  thread  ;  thus,  a  1  %"  screw, 
standard  thread,  which  is  l^s  inches  in  diameter  at  the  bottom 
of  thread,  will  be  the  bolt  to  use. 


STRENGTH    OF    MATERIALS.  2  19 

EXAMPLE  2. 

What  size  of  bolt  is  required  to  sustain  the  same  load  as  is 
mentioned   in  the  previous  example,  if  only  5  is  wanted  as  a 
factor  of  safety? 
Solution  : 

D  -  J  4450  X  5 

*  50000  X  0.7854 
D  ~  \A).5G7 

D  —  0.75  inch  diameter  in  bottom  of  thread. 
Thus  a  J^-inch  standard  screw  is  too  small,  as  that  is  only 
fij"  in  bottom  of  thread,  but  a  1-inch  standard  screw  is  suffi- 
cient, being  ||-"  in  bottom  of  thread. 

To  Find  the  Thickness  of  a  Cylinder  to  Resist  a  Given 
Pressure. 

When  the  walls  of  cylinders  are  thin  in  proportion  to  their 
diameters  use  the  formula  : 
__  S  X  / 


x  x/ 

t  _  P  X  R  X  / 


P  X/ 

P  =  Pressure  per  square  inch. 
R  =  Radius  of  cylinder  in  inches. 
/  =  Thickness  of  cylinder  wall  in  inches. 
f—  Factor  of  safety. 
S  =  Ultimate  tensile  strength  of  material. 
When  cylinder  walls  are  thick  in  proportion  to  the  diameter, 
such  as  hydraulic  cylinders,  their  thickness  is  usually  figured 
by  the  formula  : 

_      P  X   R 

s-cf>-f 

/  —  Thickness  of  cylinder  wall  in  inches. 
P  =  Pressure  in  pounds  per  square  inch. 
7?  =  Radius  of  cylinder. 
S  —  Ultimate  tensile  strength. 
f=  Factor  of  safety. 
EXAMPLE. 

^  Find  necessary  thickness  of  a  hydraulic  cylinder  of  10-inch 
inside  diameter,  made  from  cast-iron,  to  stand  a  pressure  of  1000 


22O  STRENGTH    OF    MATERIALS. 

pounds  per  square  inch,  with  4  as  factor  of  safety.    The  ulti- 
mate tensile  strength  of  the  iron  is,  by  experiments,  found  to  be 
20,000  pounds  per  square  inch.     ( See  Table  No.  25). 
Solution : 

10-inch  diameter  =  5-inch  radius. 

f  _     1000  X  5 

" 


1000  X  5 
5000  —  1000 
5000 
4000 
IX  inch. 


Strength  of  Flat  Cylinder  Heads. 

The  American  Machinist,  in  Question  No,  147,  March  22, 
1894,  gives  the  following  formula  for  flat  circular  heads  firmly 
fixed  to  the  flange  of  the  cylinder : 

/  _     /2  X  rz  X  P 
'     3  X  Si 

t  ==•  Thickness  of  cylinder  head  in  inches, 

r  =  Radius  of  cylinder  head  in  inches. 
P  ==•  Pressure  in  pounds  per  square  inch. 
Si  =  Allowable  working  stress  in  the  material. 

The  allowable  working  stress  may  be  taken  as  £  to  A  of 
the  ultimate  tensile  strength  and  may,  for  cast  iron,  be  from 
1500  to  2500  and  for  wrought  iron  from  4000  to  6000.  The 
above  formula  was  used  to  calculate  the  thickness  of  a  cast  iron 
cylinder  head  of  30  inches  diameter,  to  resist  a  pressure  of  100 
pounds  per  square  inch.  This  formula  is  in  that  case  con- 
sidered to  give  sufficient  thickness,  so  that  no  ribs  or  braces  are 
needed. 

The  above  formula  may  also  be  used  for  wrought  iron,  by 
selecting  the  proper  value  for  Si.  Assuming  the  tensile  strength 
of  wrought  iron  to  be  44,000  pounds,  and  allowing  a  factor  of 
safety  of  8,  the  value  of  Si  for  wrought  iron  will  be  5500. 

Strength  of  Dished  Cylinder  Heads. 

The  American  Machinist,  in  Question  183,  April  12,  1894, 
ves  the  following  formula  for  dished  circular  heads,  firmly 
"  to  the  flanges  of  the  cylinder : 

*     <TX  Si  xw~ 


STRENGTH    OF    MATERIALS.  221 

/  =  Thickness  of  cylinder  head  in  inches. 

R  =  Radius  of  cylinder  head  in  inches. 

P  —  Pressure  in  pounds  per  square  inch. 

d  —  Depth  in  inches  of  dishing  of  the  head  at  its  center. 

Si  =  Allowable  working  stress  in  the  material,  which  may 
be  the  same  as  given  above. 

This  formula  was  used  to  calculate  the  thickness  of  a  cast 
iron  head  44  inches  in  diameter,  dished  7  inches,  steam  pressure 
75  pounds  per  square  inch. 

NOTE.  —  In  these  examples  the  radius  of  the  bolt  circle 
should  be  considered  as  the  radius  of  the  head  when  calculat- 
ing the  thickness.  P^or  diameter  of  bolts  and  spacing  of  bolts 
see  examples  under  Steam  Engine. 

The  above  formula  may  also  be  used  for  dished  cylinder 
heads  of  wrought  iron  or  steel,  by  allowing  the  proper  value  for 
•5*1.  For  soft  steel  Si  may  be  9000  to  12,000  pounds,  and  for 
wrought  iron  5000  to  8000  pounds. 

CAUTION.  —  Cast  iron  is  not  a  desirable  material  to  use  for 
large  unribbed  cylinder  heads;  either  flat  or  dished  wrought 
iron  or  steel  is  far  superior. 

Strength  of  a  Hollow  Sphere  Exposed  to  Internal 
Pressure. 

d*     TT 

The  pressure  acts  on  a  surface  equal  to  —  -7—  and  it  is  re- 

/")      I       .V  ••• 

sisted  by  a  metal  area  equal  to  —  ^  —  X  if  X  /. 

D  —  External  diameter. 
</—  Internal  diameter. 
/  =  Thickness  of  metal. 

When  the  difference  between  inside  and  outside  diameter 
is  small  it  need  not  be  considered  in  practice,  and  the  formula 
will  be  : 


P  X  —  £-  =  dX  K  X  t  X  Si 
which  reduces  to 

p  _  4  X  /  X  Si 

d 
P  X  d 

'=-±sT 

Si  —  Allowable  tensile  stress  in  the  material. 
NOTE.  —  This  formula  only  allows  for  tensile  strength;   if  it 
is  used  for  calculating  the  thickness  of  the  body  of  a  globe 
valve  or  anything  similar  a  liberal  amount  of  metal  must  be, 
added,  in  order  to  obtain  good  results  when  casting. 


222 


STRENGTH    OF    MATERIALS. 


Strength  of  Chains. 

The  following  table  gives  approximately  the  weight  of 
wrought  iron  chains,  in  pounds  per  foot  and  kilograms  per 
meter;  and  also  their  strength,  with  six  as  factor  of  safety. 
Chains  ought  to  be  tested  with  twice  the  load  given  in  the 
table.  Never  lose  sight  of  the  fact  that  a  chain  in  use  will  wear 
and  consequently  become  reduced  in  strength;  also,  that  a 
chain  is  no  stronger  than  its  weakest  link. 


Diameter 
of   Links 
in  Inches. 

Load 
in  Pounds. 

Weight  in 
Pounds 
Per  Foot. 

Load 
in  Kilograms. 

Weight  in  Kilo- 
grams per  Meter. 

A 

280 

0.42 

125 

0.625 

1A 

500 

0.91 

225 

1.35 

H 

1125 

1.5 

510 

2.22 

1A 

2000 

2.5 

900 

3.72 

H 

4500 

5.8 

3050 

8.63 

8000 

10 

3600 

14.88 

Strength  of  Iron  Wire  Rope. 

The  following  table  gives  approximately  the  strength  of 
iron  wire  rope,  with  six  as  a  factor  of  safety. 


Diameter  in  Inches. 

Load  in  Pounds. 

Load  in  Kilograms. 

1 

1000 
2500 
3500 

453 
1134 

1588 

Wire  ropes  should  not  be  bent  over  pulleys  of  very  small 
diameter.  When  used  for  hoisting,  the  diameter  of  pulley  ought 
at  least  to  be  40  times  the  diameter  of  rope.  For  further 
information  on  wire  rope  see  manufacturers'  catalogues. 

Strength  of  Manila  Rope. 

The  size  of  manila  rope  is  measured  by  the  circumference, 
therefore  so-called  three-inch  rope  is  about  one  inch  in  diameter. 
New  manila  rope  of  three  inches  circumference  will  usually 
break  for  a  load  of  7,000  to  9,000  pounds.  For  common  use 
such  ropes  may  be  loaded  as  given  in  the  following  table,  and 
the  diameter  of  the  pulley  ought  to  be  at  least  eight  times  the 
diameter  of  the  rope. 


Size  of  Rope. 

Safe  Load  in  Pounds, 

Safe  Load  in  Kilograms. 

3  ins,  circumference, 
4    "               " 

5   "              " 

500 
800 
1300 

227 
363 
590 

STRENGTH    OF    MATERIAL. 


223 


CRUSHING    STRENGTH. 

Short  posts  having  square  ends  well  fitted  may  be 
considered  to  give  away  under  pure  crushing  stress.  Their 
strength  is  in  proportion  to  their  area;  therefore,  when  the 
length  of  a  post  does  not  exceed  four  to  five  times  its  diameter 
or  smallest  side,  its  strength  or  size  may  be  calculated  by  the 
following  formulas : 


f 


Side. of  a  square  post  =  \— ~^- 


j- 


Diameter  of  a  round  post  =  -v/- 

* 


P  =   Safe  load  in  pounds  to  be  supported  by  the  post. 
A   —  Area  of  post  in  square  inches. 

S  =   Ultimate  crushing  strength  of  the  material  in  pounds 
per  square  inch,  given  in  Table  No.  26. 
f  =   Factor  of  safety. 

TABLE   No.  26.  — Modulus  of    Elasticity  and    Ultimate 
Crushing  Strength  of  Various  Materials. 


MATERIALS. 

Modulus  of 
Elasticity 
in  Pounds 
per  Square 
Inch. 

Ultimate 
Crushing 
Strength  in 
Pounds  per 
Square  Inch 

Modulus  of 
Elasticity  in 
Kilograms 
per  Square 
Centimeter. 

Ultimate 
Crushing 
Strength  in 
Kilograms  per 
Sq.  Centimeter 

Cast  steel   .... 
Bessemer  steel  .    . 

Wrought  iron    .   < 

Cast  iron    .    .    .   ) 

Oak          (endwise) 
Pitch  pine      " 
Pine 
Spruce            " 

T,     •     1                                                     ( 

30,000,000 
28,000,000 

25,000,000 

12,000,000 
to 
15,000,000 
1,500,000 
1,600,000 
1,100,000 
11,000,000 

150,000 
50,000 
45,000 
to 
50,000 

90,000 

9,000 
9,000 
6,000 
6,000 
800 

2,200,000 
1,970,000 

1,700,000 

800,000 
to 
1,000,000 
105,000 
112,000 
77,000 
77,000 

10,500 
3,500 
3,000 
to 
3,500 

6,300 

630 
630 
420 
420 
56 

Brick   < 

Brick   work  laid  i 
in  1  part  cement  > 

to    2,000 
600 

to       140 
42 

and  3  parts  sand  ) 
Brick  work  laid  ) 

240 

16  to  17 

in  lime  and  sand  ] 
Granite    . 

10,000 

700 

224 


STRENGTH    OF   MATERIALS. 


FIG.  1 


When  a  post  or  column  is  long  compared  to  its  diameter, 
its  strength  will  decrease  as  the  length  is  increased.  Anyone 
will,  from  every-day  observation,  know  that  a  short  post  will 
support  with  perfect  safety  a  load  which  will  break  a  long  one. 

Short  columns  break  under  crushing,  but  long  ones  break 
under  comparatively  light  load  by  the  combined  effect  due  to 
both  crushing  and  flexure.  It  is,  therefore,  evident  that  the 
strength  of  long  columns  follows  laws  very  different  from  those 
which  apply  to  short  ones. 

The  form  of  the  ends  has  also 
great  influence  on  the  strength  of  a 
column  when  under  crushing  and 
deflective  stress.  (  See  Fig.  1). 

When  both  ends  are  round  the 
column  has  least  strength;  if  one 
end  is  round  and  one  end  flat  it  is 
stronger,  but  if  both  ends  are  flat 
and  square  with  the  center-line,  it  is 
strongest.  The  proportions  are  ap- 
proximately as  1,  2  and  3. 

Eccentric  loading  on  columns  will  also  have  a  very  destruct- 
ive effect  upon  their  strength. 

Theoretical  calculations  regarding  the  strength  of  columns 
and  posts  are  difficult,  and  such  empirical  formulas  as  the 
well-known  Hodgkingson's  or  Gordon's  formulas  are  usually  re- 
sorted to. 

The  Hodgkingson  formulas  for  long  columns  having  square 
ends  well  fitted  are : 


P  =  99,000   X 


for  solid  cast  iron  columns. 


P  =  99,000  X  D  °  ~,f       for  hollow  cast  iron  columns. 
J~* 

P  =  285,000  X  D  \2     for  solid  wrought  iron  columns. 

P  =  Breaking  load  in  pounds. 
D  =  External  diameter  in  inches. 
d=  Internal  diameter  in  inches. 
L  =  Length  in  feet. 

When  the  breaking  load  as  calculated  by  these  formulas 
exceeds  one  quarter  of  the  crushing  load  of  a  short  column  of 
the  same  metal  area,  the  result  must  be  corrected  by  the  for- 
mula; 


STRENGTH  OF  MATERIALS. 
PX  C 


225 


-  P  4-  %  C  X  A 

P\  =  Corrected  breaking  load  of  column. 
C=  Crushing  strength  of  material  (see  Table  No.  26). 
A  =  Metal  area  of  column  in  square  inches. 
IMPORTANT. —  Applying  the  last  formula,  the  result,  / 
must  always  be  smaller  than  /'. 

Table  No.  27  was  calculated  by  the  following  formulas : 

(  36000 

Column  I.        Safe  Load  =  0.1  X  <  ~       ~fz —  ~  x 

(1     -f-    (=T2-    X    0.00025) 

36000  X  0.07031 
+(~  X  0.00025) 


Column  II.      Safe  Load  =  0.1  X 


36000 


Column  III.     Safe  Load  =  0.1  X 


Column  IV.     Safe  Load  =  0.1  X 


1  +  /—  X  0.0004 
v/3* 

36000  X  0.07031 


0.0004) 


80000 


Column  V.        Safe  Load  =  0.1  X    j  1  -f  t—  X  0.0025  ) 

80000  X  0.07031 
Column  VI.      Safe  Load  =  0.1  X     )  1  +  (  —  X  0.0025  ^ 

(  80000 


Column  VII.    Safe  Load  =  0.1  X 


Column  VIII.  Safe  Load  —  0.1  X 


Column  IX.      Safe  Load  =  0.1  X 


Column  X.        Safe  Load  =  0.1  X 


1  +  (  ±_  x  0.0035  ) 

80000  X  0.07031 
1  +(^X  0.0035) 

5000 

I  1  +  (^X  0.004) 
5000  X  0.07031 


{ 


-,     , 
]     h 


Z2 


226 


STRENGTH   OF  MATERIALS. 


TABLE  No.    27.— Safe  Load    on  Pillars  Having  Square 
Ends  Well  Fitted. 

(10  is  used  as  Factor  of  Safety). 


U 

WROUGHT  IRON. 

CAST  IRON. 

WOOD. 

w 

m 

1. 

Hollow  Pillar 

Solid  Pillar. 

Hollow  Pillar. 

Solid  Pillar. 

[opruce  or 
white  Pine) 

3 
.2  v 

'•3-S 
xs 

| 

| 

f 

<j 
a 

!• 

1 

| 

h* 

fi 

V 

^ 
u 

8 

i 

*c 

o 

JU 

"S 

T3-O 
£« 

*O  *) 

11 

V 

j-i  OS 

H 

JH  CS 

&| 

&c 

u 

i-  rt 

-  S 

&a 

« 

1*  03 

M 

e 

IH  rt 

H 

"O  I) 
|| 

IS 

•£  ° 

ti 

Ki 

f 

13 

V  3 

rl 

13 
B 

g3 
«  v 

5P« 

Jl 

c 

g<3 

2  jj 

V  3 
|* 

iu 

^ 

If 
</>'S> 

T3 

a 

§3 

2  « 

MS 

'•3  "= 
•5  ° 

60 

3 

1 

W. 

1 

11 

1 

ll 

1 

°.  3 
££ 

I 

sal 

St? 

3 

L 

£ 

~D~ 

i. 

ii. 

III. 

IV. 

V. 

VI. 

VII. 

VIII. 

IX. 

X. 

~D~ 

4 

3585 

252 

3567 

252 

7692 

540 

7624 

536 

476 

34 

4 

Q 

3567 

251 

3549 

250 

7339 

512 

7105 

500 

438 

30 

6 

8 

3545 

250 

3510 

247 

6896 

484 

6536 

460 

398 

28 

8 

10 

3512 

247 

3462 

243 

6400 

450 

5926 

417 

357 

25 

10 

12 

3475 

244 

3404 

239 

5882 

413 

5314 

374 

317 

22 

12 

14 

3432 

241 

3338 

235 

5369 

371 

4745 

333 

280 

20 

14 

16 

3383 

238 

3266 

230 

4878 

343 

4226 

297 

248 

17 

16 

18 

3332 

234 

3187 

224 

4420 

311 

3749 

264 

218 

15 

18 

20 

3273 

230 

3103 

218 

4000 

281 

3333 

234 

192 

13 

20 

22 

3211 

226 

3018 

212 

3620 

255 

2969 

208 

170 

12 

22 

24 

3147 

221 

2926 

206 

3279 

230 

2653 

186 

151 

10.6 

24 

26 

3080 

216 

2834 

199 

2974 

211 

2376 

167 

135 

9.5 

26 

28 

3010 

211 

2741 

192 

2703 

190 

2137 

150 

121 

8.5 

28 

30 

2938 

206 

2647 

186 

2462 

173 

1928 

136 

109 

7.6 

30 

32 

2866 

201 

2554 

180 

2247 

158 

1745 

123 

98 

6.9 

32 

34 

2793 

196 

2462 

173 

2056 

145 

1585 

111 

88 

6.2 

34 

36 

2719 

191 

2370 

167 

1887 

133 

1440 

102 

81 

5.7 

30 

38 

2645 

186 

2288 

160 

1735 

122 

1321 

93 

74 

5.2 

38 

40 

2571 

181 

2195 

154 

1600 

112 

1212 

85 

67 

4.7 

40 

42 

2498 

176 

2111 

148 

1479 

104 

1113 

78 

62 

4.3 

42 

44 

2426 

171 

2029 

143 

1370 

96 

1028 

74 

57 

4 

44 

46 

2354 

166 

1950 

137 

1272 

89 

952 

67 

53 

3.8 

46 

48 

2284 

161 

1874 

132 

1183 

83 

882 

62 

49 

3.4 

48 

50 

2215 

155 

1800 

127 

1103 

78 

820 

58 

45 

3.2 

50 

This  table  is  intended,  in  connection  with  Table  No.  24,  to 
facilitate  calculations  for  pillars  of  either  wood  or  iron,  and  may 
be  used  with  equal  advantage  for  English  or  metric  measures, 
provided  both  diameter  and  length  are  taken  by  the  same  system. 

For  a  round  pillar  divide  the  length  by  the. diameter,  but 
for  a  square  or  rectangular  pillar  divide  the  length  by  the 


STRENGTH   OF   MATERIALS.  22  ^ 

smallest  side.  Find  the  quotient  in  the  column  headed  -p- 
and  find  the  safe  load  per  square  unit  of  area  in  the  cor- 
responding column  of  the  table.  Multiply  this  by  the  metal 
area  of  the  pillar,  and  the  product  is  the  safe  load,  with  10  as 
factor  of  safety,  on  a  pillar  well  fitted  and  having  square  ends. 
For  any  other  kind  of  ends  and  any  other  unit  of  safety,  allow- 
ance must  be  made  as  explained  on  previous  pages. 

EXAMPLE  1. 

Find  the  safe  load  in  pounds,  according  to  Table  No.  27, 
for  a  round,  hollow,  cast-iron  pillar  five  feet  long,  five  inches 
outside  and  four  inches  inside  diameter,  having  square  ends 
well  fitted  and  being  evenly  loaded. 

Solution : 

Five  feet  equals  60  inches,  and  00  divided  by  5  gives  12. 
In  the  first  column,  under  the  heading  "  Length  divided  by 
diameter,  or  smallest  side, "  is  12,  and  in  that  line,  in  the  column 
headed  **  pounds  per  square  inch  "  for  hollow  cast-iron  pillars, 
is  5882.  The  metal  area  of  this  pillar  is  obtained  by  subtract- 
ing the  area  of  a  circle  four  inches  in  diameter  from  the  area  of 
a  circle  five  inches  in  diameter  (see  area  of  circles,  page  196; 
Table,  page  209),  which  is  19.68  —12.57  =  7.06,  or  practically 
seven  square  inches,  and  seven  times  5882  equals  41,174  pounds. 

EXAMPLE  2. 

Find  the  safe  load  in  kilograms,  according  to  Table  No.  27, 
for  a  round  spruce  post  2  meters  long  and  20  centimeters  in 
diameter. 

Solution : 

Two  meters  =  200  centimeters  and  2^s°  —  10.  The  corre- 
sponding constant  in  the  table  is  25  kilograms.  The  area  of  a 
circle  20  centimeters  in  diameter  is  814.2  square  centimeters, 
and  25  times  314.2  =•  7855  kilograms,  as  safe  load. 

EXAMPLE  3. 

What  would  be  the  safe  load  on  the  same  post  if  it  had 
been  20  centimeters  square,  instead  of  round  ? 

Solution : 

The  length  is  the  same;  therefore  the  length  divided  by  the 
side  gives  10,  as  before,  and  the  corresponding  constant  is  25 
kilograms,  but  as  the  cross-sectional  area  in  square  centimeters 
is  20  X  20  =  400,  the  corresponding  load  will  be  400  X  25  = 
10,000  kilograms  as  the  safe  load. 

NOTE. — It  will  be  noticed  that  in  figuring  the  strength  of 
pillars  according  to  this  table,  the  strength  of  a  square  pillar 
will  always  be  to  the  strength  of  a  round  pillar  as  1  to  0.7854, 
while  theoretically  the  strength  of  a  square  pillar  compared  to 
that  of  a  round  pillar  will  vary  with  the  length,  the  extremes 
being  1  to  0.589  for  extremely  long  pillars  and  1  to  0.7854  for 


228  STRENGTH   OF   MATERIALS. 

very  short  ones.  This  discrepancy  is  frequently  unimportant  in 
practical  work,  because  pillars  are  usually  comparatively  short, 
and  also  because  a  high  factor  of  safety  is  always  used,  but  it 
is  well  to  remember  and  provide  for  this  fact  in  cases  of  very 
long  pillars. 

Hollow  Cast=Iron  Pillars. 

By  referring  to  the  formulas  and  considering  the  laws 
governing  the  strength  of  pillars,  it  is  seen  that  the  strength 
of  pillars  increases  very  fast  by  increasing  their  diameter  or 
their  sides.  In  cast-iron  pillars  this  is  taken  advantage  of  by 
making  them  large  in  diameter  and  coring  out  the  stock  on  the 
inside. 

The  thickness  of  the  metal  may  be  about  TV  of  the  diameter 
of  the  pillar.  In  small  pillars  it  must  be  thicker  in  order  to 
obtain  good  results  when  casting.  A  flange  is  cast  on  each  end 
to  form  enough  bearing  surface,  and  the  pillar  is  squared  off 
very  carefully  so  that  both  ends  are  square  with  the  center-line. 
This  is  an  important  point,  as  the  strength  is  enormously 
destroyed  by  squaring  the  ends  carelessly  and  thereby  bringing 
the  load  to  act  corner-ways  on  the  pillar. 

Table  No.  28  was  calculated  by  the  formula : 
80000  X  metal  area   >> 


(  J 

Safe  Load  =  0.1  X    j  1 


+  (^X  0.0025) 


and  the  result  obtained  reduced  to  long  tons  (2240  pounds). 
Ten  is  thus  used  as  a  factor  of  safety ;  both  ends  of  the  pillar 
are  supposed  to  be  square  and  evenly  loaded.  For  other  shapes 
of  ends,  mode  of  loading,  or  other  factors  of  safety,  propor- 
tional allowance  must  be  made.  For  instance,  if  15  is  required 
as  factor  of  safety,  allow  only  two-thirds  of  the  load  given  in 
the  table. 

If  the  pillar  has  only  one  square  end  and  one  round  end, 
allow  only  two-thirds  as  much  load.  If  it  has  both  ends 
rounded,  or,  which  is  the  same,  if  the  ends  have  only  a  very 
imperfect  bearing,  allow  only  one-third  as  much  load. 

Weight  of  Cast=Iron  Pillars. 

The  weight  of  a  cast-iron  pillar  may  be  calculated  by  the 
formula : 

W  —  (D*  —  d'2}  X  L  X  2.45 
W  =  Weight  of  pillar  in  pounds. 
D  =  Outside  diameter  in  inches. 
d  —  Inside  diameter  in  inches. 
L  =  Length  of  pillar  in  feet. 

The  weight  given  in  Table  No.  28  was  calculated  by  this 
formula,  and  the  length  taken  as  one  foot. 


STRENGTH   OF   MATERIALS.  2 29 

TABLE  No.  28.— Safe  Load  on  Round  Cast-Iron  Pillars. 


11 

•-S  (J 

ih  * 

e  5 

'"  O.C.C 

Length  of  Pillars  in  Feet. 

External  c 
eter  in  in 

Thicknei 
Metal  in  it 

<  <j 

•3-S.g 

1  s 

fiif 

fir* 

1 

«o 

I 

£ 

10  Feet. 

12  Feet. 

I 

^< 

1 

CD 

1 

eo 

I 

8 

1 

8 

1 
s 

4 

# 

5.49 

17.14'  11 

8.1 

6.1 

4 

y* 

7.56 

23.90 

15.2 

11.3 

8.5 

5 

% 

7.07 

22.06 

16.8 

13.3 

10.4 

8.3 

5 

X 

10.01 

31.23 

24 

19 

15 

12 

0 

l/2 

8.64 

26.95 

23 

19 

15.5 

12.6 

10.4 

0 

H 

12.37 

38.59 

33 

27 

22 

18 

15 

6 

H 

14.09 

43.96 

37 

31 

25 

21 

17 

6 

i 

15.71 

49.01 

42 

35 

28 

23 

19 

6 

1/8 

17.23 

53.76 

47 

40 

32 

26 

22 

7 

* 

12.52 

39.06 

36 

31 

26 

22 

19 

10 

7 

X 

14.73 

45.96 

42 

36 

31 

26 

22 

19 

7 

H 

16.84 

52.54 

48 

41 

35 

29 

25 

21 

7 

i 

18.85 

58.90 

54 

46 

39 

33 

29 

24 

7 

i# 

20.76 

64.77 

60 

52 

44 

37 

32 

27 

8 

X 

17-08 

53.29 

51 

45 

39 

34 

29 

25 

22 

8 

H  119.59 

61.12 

59 

52 

45 

39 

34 

29 

25 

8 

1 

21.99 

68.64 

66 

58 

51 

44 

38 

33 

28 

8 

1# 

24.30 

75.82 

73 

64 

56 

48 

42 

36 

31 

8 

IX 

26.51 

82.71 

79 

70 

61 

52 

45 

39 

34 

8 

1/8 

28.62 

89.29 

86 

76 

66 

57 

48 

42 

37 

9 

X 

19.44 

60.65 

60 

54 

49 

43 

37 

33 

29 

24 

1) 

H 

22.33 

69.67 

69 

63 

56 

49 

43 

38 

33 

29 

9 

1 

22.13 

78.40 

78 

71 

63 

55 

48 

42 

37 

33 

9 

1* 

27.83 

86.83 

87 

78 

69 

62 

53 

47 

41 

36 

9 

IX 

30.43 

94.94 

95 

85 

76 

67 

58 

51 

45 

39 

9 

1/8 

32.94 

102.77 

102 

92 

82 

72 

63 

55 

48 

43 

9 

1# 

35.34 

110.26 

110 

99 

88 

78 

68 

59 

52 

46 

9 

1% 

39.86 

124.36 

126 

113 

100 

90 

78 

67 

60 

51 

10 

H 

25.09 

78.28 

80 

73 

67 

60 

53 

47 

42 

37 

34 

10 

1 

28.28 

88.23 

90 

83 

75 

67 

60 

53 

47 

42 

38 

10 

i# 

31.37 

97.87 

100 

92 

83 

74 

66 

58 

52 

47 

42 

10 

IX 

34.37 

107.23 

110 

101 

91 

82 

73 

64 

57 

51 

47 

10 

1/8 

37.26 

116.25 

119 

109 

98 

88 

79 

69 

62 

55 

51 

10 

i# 

40.06 

124.99 

128 

117 

106 

95 

85 

75 

67 

59 

54 

10 

IX 

45.36 

141.52 

146 

133 

122 

109 

97 

85 

77 

67 

60 

11 

1 

31.42 

98.04 

102 

95 

87 

79 

71 

64 

58 

52 

48 

43 

11 

IH 

34.90 

108.89 

114 

105 

96 

88 

79 

71 

64 

58 

53 

48 

11 

IX 

38.29 

119.46 

125 

116 

106 

97 

87 

78 

70 

63 

58 

52 

11 

IK 

41.58 

129.73 

135 

126 

115 

105 

94 

85 

76 

68 

62 

56 

11 

i# 

44.77 

139.68 

146 

136 

124 

113 

102 

92 

82 

74 

68 

61 

11 

IX 

50.86 

158.68 

166 

156 

142 

129 

178 

106 

94 

86 

79, 

71 

11 

2 

56.55?  176.44 

186 

176 

160 

147 

134 

120 

106 

98 

90 

81 

230 


STRENGTH    OF   MATERIALS. 


TABLE  No.  28.  — (Continued). 


11 

2.s 

c.S 

gj  l-i 

ickness  of 
al  in  inches 

etal  Area 
in 
[.  inches. 

.5  « 

fill 

t^-  o  *™3 

Length  of  Pillars  in  Feet. 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

£  v 

£" 

El 

&  d 

^P-t 

FBI 

CO 

MH 
OO 

0 

23 

«0 

oo 

§ 

S 

3 

12 

1 

34.46 

107.51 

115 

107 

99 

92 

83 

^76 

68 

62 

5S|  53 

12 

\y% 

38.34 

119.62 

128 

119 

108 

102 

92 

84 

78 

69 

63  58 

12 

IX 

42.12 

131.41 

141 

131 

119 

112 

101 

93 

84 

76 

70  04 

12 

\y% 

45.80 

142.90 

153 

142 

129 

121 

110 

101 

91 

82 

75  69 

12 

IX 

49.39 

154.10 

165 

154 

139 

131 

119 

109 

99 

89 

82 

75 

12 

IX 

56.26 

175.53 

189 

178 

159 

150 

137 

125 

115 

103 

94 

85 

12 

2 

62.74 

195.75 

213 

201 

179 

170 

155 

141 

131 

117 

106 

96 

13 

1 

37.97 

117.53 

127 

119 

111 

104 

97 

87 

79 

73 

67 

61 

13 

\y% 

41.94 

130.85 

142 

134 

126 

117 

109 

101 

91 

82 

75 

69 

13 

IX 

46.11 

143.86 

158 

149 

140 

130 

121 

112 

101 

91 

84 

77 

13 

50.19 

156.59 

174 

163 

154 

144 

133 

122 

111 

101 

93 

84 

13 

1^ 

54.16 

168.98 

190 

178 

168 

157 

145 

133 

121 

110 

101 

92 

13 

IX 

61.82 

192.88 

214 

201 

189 

176 

164 

151 

137 

124 

114 

104 

13 

2 

69.09 

215.56 

227 

224 

210 

195 

182 

168 

152 

137 

126 

116 

14 

1 

40.94 

127.60 

138 

131 

123 

115 

109 

101 

92 

85 

78 

72 

14 

IX 

45.50 

141.96 

153 

145 

139 

130 

121 

112 

103 

94 

87 

80 

14 

IX 

50.07 

156.31 

168 

160 

153 

143 

133 

123 

113 

104 

95 

88 

14 

54.54 

170.04 

183 

174 

167 

156 

145 

134 

123 

113 

104 

95 

14 

\y2 

58.78 

183.67 

198 

189 

180 

168 

156 

145 

133 

122 

112 

103 

14 

IX 

67.31 

210.00 

226 

216 

206 

192 

179 

166 

152 

140 

128 

118 

14 

2 

75.36 

235.12 

254 

242 

232 

216 

201 

186 

171 

157 

143 

133 

15 

1 

43.99 

137.28 

150 

143 

136 

128 

120112 

104 

96 

90 

82 

15 

\y^ 

49.04 

153.19 

167 

159 

152 

143 

136  125 

116 

108 

100 

93 

15 

IX 

54.00 

168.48 

184 

175 

167 

157 

148138 

128119 

110 

102 

15 

58.85 

183.77 

201 

191 

182 

172 

161  151 

140 

130 

120 

111 

15 

\y2 

63.62 

198.74 

217 

207 

197 

186 

175 

163 

151 

141 

130 

120 

15 

IX 

72.85 

227.45 

284 

236 

225 

212 

202 

190  175 

160 

148 

137 

15 
16 

2 
1 

81.68 
47.12 

254.81 
146.95 

279 
160 

265 
156 

253 
150 

237 
140 

2292161971179 
133  125  117  110 

166 
101 

154 
93 

16 

\y^ 

52.57 

164.11 

179 

173 

167 

158 

148  139  130  122 

116 

106 

16 

IX 

57.92 

180.65 

198 

191 

185 

174 

163  153  143 

135 

128 

117 

16 

63.18 

197.18 

216 

209 

202 

190 

178  167 

152 

147 

139 

128 

16 

l/^ 

68.33 

213.10 

234 

227 

220 

206 

193  181 

170 

160 

151 

138 

16 

IX 

78.34 

244.29 

268 

258 

250 

237 

224  207 

199 

183 

173 

158 

16 

2 

87.96 

274.56 

301 

290 

281 

267 

255 

233!218 

206 

194 

177 

The  length  of  cast-iron  pillars,  as  a  rule,  ought  not  to  ex- 
ceed 20  to  25  times  their  diameter.  Cast-iron  pillars,  when 
heavily  loaded,  are  apt  to  be  broken  if  struck  by  a  blow 
sidewise. 


STRENGTH    OF   MATERIALS. 


Wrought  Iron  Pillars. 

In  important  work,  cast-iron  pillars  are  rapidly  going  out  of 
use.  Wrought  iron  pillars  are  now  made  which  compare 
favorably  in  price  and  are  far  more  reliable  than  those  of  cast 
iron.  For  full  information  regarding  weight  and  strength  of 
wrought  iron  pillars  and  Z  bar  columns,  see  manufacturers' 
catalogues. 

Wooden  Posts. 

Table  No.  29  was  calculated  by  the  formula : 
, 5000  X  Area 

Safe  load  =  0.1  X    < 


nj  x  /\rea         -\ 
-j^-X  0.004  )  \ 


and  the  result  divided  by  2240. 

L  =  Length  of  post  in  inches. 

D  =  Side  of  post  in  inches. 

TABLE  No.  29.— Safe  Load  in  Tons  on  Square  Pine  or 
Spruce  Posts  Having  Square  Ends  Well  Fitted. 

(10  as  Factor  of  Safety). 


!J 

Side  of  Post  in  Inches. 

1  in. 

2  in. 

3  in. 

4  in. 

5  in. 

6  in. 

7  in. 

8  in. 

9  in. 

10  in. 

12  in. 

i 

2 
3 
4 
5 

6 
7 
8 
10 
12 
14 
16 
18 
20 
22 
24 
26 
28 
30 
32 
34 
36 
38 
40 

0.14 
0.06 
0.03 

0.78 
0.56 
0.39 
0.32 
0.19 
0.14 

1.89 
1.59 
1.27 
0.99 
0.77 
0.60 
0.49 
0.39 
0.27 

3.45 
3.12 
2.70 
2.27 
1.8S 
1.56 
1.29 
1.08 
0.78 
0.58 
0.44 

5.00 
5.00 
4.62 
4.08 
3.54 
3.05 
2.62 
2.25 
1.69 
1.29 
1.00 
0.90 
0.80 

8.00 

8.00 
7.00 
6.38 
5.74 
5.00 
4.50 
4.00 
3.09 
2.43 
1.04 
1.58 
1.30 
1.09 
0.90 

10.00 
10.00 
10.00 
9.21 
8.45 
7.68 
6.94 
6.24 
5.03 
4.06 
3.31 
2.73 
2.27 
1.92 
1.63 
1.41 
1.22 

12.50 
12.50 
12.50 
12.50 
11.21 
10.80 
9.91 
9.08 
7.53 
6.25 
5.19 
4.34 
3.66 
3.11 
2.66 
2.31 
2.01 
1.78 
1.56 

16.00 

16.00 
16.00 
16.00 
15.35 
14.35 
13.41 
12.44 
10.58 
8.96 
7.57 
6.43 
5.49 
4.72 
4.17 
3.55 
3.11 
2.68 
2.38 
2.18 
1.96 

20.00 
'20.00 
20.00 
20.00 
19.51 
18.44 
17.40 
16.32 
14.18 
12.22 
10.50 
9.04 
7.80 
6.76 
5.89 
5.17 
4.56 
4.05 
3.56 
3.23 
2.91 
2.64 
2.39 

30.00 

30.00 
30.00 

30.00 
'29.22 
28.10 
26.87 
25.50 
22.95 
20.40 
18.02 
15.88 
14.00 
12.36 
10.95 
9.73 
8.68 
7.77 
6.99 
6.30 
5.71 
5.18 
4.74 
4.35 

•    • 

-  • 

•  • 

•  • 

•  • 

232  STRENGTH    OF   MATERIALS. 

The  preceding  Table  gives  the  safe  load  in  long  tons  corre- 
sponding to  a  square  post  of  the  dimensions  of  sides  given  at 
top  of  the  columns,  and  lengths  given  in  the  first  column.  For 
round  posts  the  load  should  be  0.75  to  0.6  of  the  given  load  de- 
pending upon  the  length  of  post. 

EXAMPLE. 

What  size  of  post  is  required,  with  10  as  factor  of  safety, 
to  support  a  load  of  five  tons,  when  the  length  of  the  post  is  16 
feet? 

Solution : 

In  the  column  headed  "Length  of  post  in  feet"  find  10, 
and  in  line  with  16  find  the  numbers  nearest  to  five  tons,  which 
are  4.34  and  6.43.  Thus,  a  post  16  feet  long  and  8  inches  square 
will  support  4.34  tons,  and  a  post  16  feet  long  and  9  inches 
square  will  support  6.43  tons.  It  is,  therefore,  best  to  select  a 
post  9  inches  square. 


To  Calculate   the   Strength   of  Rectangular    Posts    from 
the  Table. 

Find,  in  the  Table,  the  strength  of  the  post  according  to 
its  smallest  side,  and  increase  the  tabular  value  in  proportion 
to  the  largest  side  of  the  post. 

EXAMPLE. 

What  is  the  strength,  with  10  as  factor  of  safety,  of  a  spruce 
post  10  feet  long,  6  inches  thick,  8}4  inches  wide,  with  square 
ends  well  fitted,  calculated  by  Table  No.  29. 

Solution : 

In  the  Table  we  find  the  strength  of  a  post  10  feet  long 
and  6  inches  square  to  be  3.09  tons.  Therefore,  when  the  pillar 
is  6  inches  thick  and  8%  inches  wide  its  corresponding  strength 

will  be  3.09  X  ^  =  4.38  tons. 

It  is  a  waste  of  material  to  use  a  post  of  rectangular  cross- 
section.  For  example,  this  post  is  6  X  8%  inches  =  51  square 
inches  of  cross-section  and  will  support  4.38  tons,  but  a  post  of 
the  same  length  and  7X7  inches  =  49  square  inches  of  cross- 
section,  will  support  5.03  tons.  (  See  Table  No.  29). 


To  Obtain  the  Weight  of  Pillars  in  Kilograms  per  Meter 
when  the  Weight  in  Pounds  per  Foot  is  Known. 

Multiply  the  weight  in  pounds  per  foot  by  the    constant 
1.4882,  and  the  product  is  the  weight  in  kilograms  per  meter. 


STRENGTH  OF  MATERIALS.  233 


TRANSVERSE    STRENGTH. 

A  beam  placed  in  a  horizontal  position,  fastened  at  one 
end  and  loaded  at  the  other,  is  exposed  to  transverse  stress, 
and  will  usually  bend  more  or  less,  as  shown  (exaggerated)  in 
Fig.  1,  before  it  will  break.  The  line 
F|G-  *•  a  b  is  called  the  neutral  line,  and  all 

fibres  above  the  neutral  line  are  exposed 
to  tensile  stress,  and  all  fibers  below 
are  exposed  to  crushing  stress,  but  the 
neutral  fiber  is  neither  stretched  nor 
compressed.  A  line  drawn  in  a  hori- 
zontal direction,  at  right  angles  to,  and 
through  the  neutral  line,  is  called  the 
neutral  axis  with  reference  to  this  particular  place  of  the  section 
of  the  beam.  The  neutral  axis  is  considered  to  pass  through 
the  center  of  gravity  of  the  section,  which,  for  beams  of  round, 
square  or  rectangular  section,  is  always  in  the  geometrical  cen- 
ter. Therefore,  all  beams  of  such  section  will  have  an  equal 
amount  of  material  on  the  upper  and  under  side  of  the  neutral 
axis,  but  it  is  not  always  desirable  for  all  materials  or  for  all 
kinds  of  load  to  have  an  equal  amount  of  material  on  both  the 
side  exposed  to  compression  and  that  exposed  to  tension.  For 
instance,  cast-iron  beams  are  usually  made  in  T  formed  sec- 
tion and  should  always  be  laid  so  that  the  largest  web  is 
exposed  ^  to  tensile  stress,  because  cast-iron  offers  much 
more  resistance  to  compression  than  it  does  to  tension.  Cast- 
iron  beams  of  such  section  ought,  therefore,  to  be  laid  in  this 
position  (T),  if  fastened  at  one  end  and  loaded  at  the  other, 
but  should  be  laid  in  this  position  ( -X. ),  if  they  are  supported 
under  both  ends  and  loaded  between  the  supports.  If 
this  is  taken  into  consideration  in  placing  a  cast-iron  beam,  its 
ultimate  transverse  breaking  strength  is  greatly  increased,  but 
under  a  moderate  load  the  deflection  will  be  practically  equal 
in  either  position,  because  as  long  as  the  load  is  small,  well 
within  the  elastic  limit,  cast-iron  will  stretch  under  tensile  stress 
as  much  as  it  will  compress  under  an  equal  amount  of  crushing 
stress;  therefore,  the  modulus  of  elasticity  for  tension  and 
compression  of  cast-iron  is  considered  to  be  equal,  but  under 
increased  crushing  load  the  compression  becomes  less  in  pro- 
portion to  the  load  until  the  point  is  reached  when  the  cast-iron 
can  not  compress  more,  and  the  casting  will  break.  The 
ultimate  crushing  strength  of  cast-iron  is  five  to  six  times  as 
much  as  its  ultimate  tensile  strength. 

A  beam  supported  under  both  ends  and  loaded  in  the 
middle  will  carry  four  times  as  great  a  load  as  another  beam  of 
the  same  size  and  material  fixed  at  one  end  and  loaded  at  the 


234 


STRENGTH    OF    MATERIALS. 


251bs.        FIG.   2. 


100  Ibs. 


ft.  J. 2  feet_._ 


FIG.  3. 


50  Ibs. 


other.     This  may  be  understood  by  .  l  foot. 

referring  to  Fig.  2,  as  when  the  beam  , 

is  one  foot  long   and  loaded  with  4~ 

100  pounds  in  the  middle,  each  half  /—A 
of  the  beam  supports  only  50  pounds, 
and  this  50  pounds  acts  only  upon  an 
arm  y2  foot  long,  consequently  it 
exerts  no  more  force  toward  break- 
ing this  beam  than  the  25  pounds 
would  upon  the  end  of  the  other 
beam  one  foot  long. 

A  beam  twice  as  wide  as  another  and  of  the  same  length, 
thickness,  and  material,  will  carry  twice  the  load,  because  the 
wide  beam  could,  of  course,  be  split  into  two  equal  beams  ; 
consequently  it  must,  as  a  whole 
beam,  have  twice  the  strength  of 
another  one  of  the  same  material 
but  of  only  half  the  width. 

A  beam  twice  as  long  as  another 

will  break  under  half  the  load.  This       / 

is  seen  by  referring  to  Fig.  3,  be-        looibs. 

cause  50  pounds  on  an  arm  two  feet 

long  will  balance  100  pounds  on  an  arm  one  foot  long. 

A  beam  twice  as  thick  as  another,  of  the  same  material, 
length  and  width,  will  carry  four  times  the  load.     ( See  Fig.  4). 

Suppose  the  weight  a  is  act- 
ing on  the  arm  b,  tending  to  swing 
it  around  the  center  c,  and  this 
action  being  counteracted  by  the 
weights  g  and  h,  also  by  the 
arrows  e  and  /.  If  the  weight  h 
is  taking  hold  twice  as  far  from 
the  center  as  the  weighty  it  will 
offer  twice  the  resistance  against 
swinging  the  beam  that  ^will; 
and  exactly  the  same  with  the 

arrowsyand  e.  Consider  the  line  c  b  as  the  neutral  fiber,  the 
arrows  e  and  f  as  representing  the  fibers  resisting  crushing, 
and  the  weigh  ts^and  h  as  representing  the  fibers  resisting  tensile 
stress.  It  will  be  understood  that  if  the  fibers  are  twice  as  far 
above  or  below  the  neutral  fiber  they  are  in  a  position  to  offer 
twice  the  resistance  to  the  breaking  action  of  the  load  ;  but  a 
beam  of  twice  the  thickness  has  not  only  its  average  fiber  twice 
as  far  from  the  neutral  point,  but  it  has  also  twice  the  area  or 
twice  as  many  fibers,  consequently  the  result  must  be  that  it 
can  resist  four  times  the  load, 


FIG.  4. 


O«. 


STRENGTH    OF  MATERIALS. 


235 


For  instance  :  The  beam  a  in  Fig- 
ure 5  is  four  times  as  strong  as  the 
beam  b,  if  placed  on  the  edge,  as  shown 
in  the  figure,  and  loaded  on  the  top  ; 
but  a  would  be  only  twice  as  strong  as 
b  if  it  was  laid  on  the  side  and  loaded 
on  top. 


FIG.  5. 


•4in. 


Formulas  and  Rules  for  Calculating  Transverse  Strength 
of  Beams. 

The  fundamental  formula  for  transverse  stress  in  beams  is: 
BENDING  MOMENT  =  RESISTING  MOMENT. 

The  bending  moment  for  a  beam  fixed  at  one  end  and 
loaded  at  the  other  (see  Fig.  1)  is  obtained  by  multiplying  the 
load  by  the  horizontal  distance  from  the  neutral  axis  to  the 
point  where  the  load  is  applied.  The  distance  is  taken  in 
inches  and  the  load  in  pounds. 

The  resisting  moment  is  obtained  by  multiplying  the  mo- 
ment of  inertia  by  the  unit  stress,  tensile  or  compressive,  upon 
the  fiber  most  remote  from  the  neutral  axis,  and  dividing  the 
product  by  the  distance  from  this  fiber  to  the  neutral  axis. 

The  theoreticarformula  for  the  transverse  strength  of  a 
beam  fastened  in  a  horizontal  position  at  one  end  and  loaded  at 
the  extremity  of  the  other  end,  as  shown  in  Fig.  C,  is, 


L  X  a 

When  the  beam  is  fastened  at  one  end  and  loaded  evenly 
throughout  its  whole  length,  as  shown  in  Fig.  7,  the  formula 
will  be, 

C  V   V 

P  =  2  X  °  x  l 
L  X  a 

When  a  beam  is  placed  in  a  horizontal  position  and  sup- 
ported under  both  ends  and  loaded  in  the  middle  (see  Fig.  8) 
the  formula  is, 

s  x  r 

L  X  a 

When  a  beam  is  placed  in  a  horizontal  position  and  sup- 
ported under  both  ends  and  loaded  throughout  its  whole  length 
(see  fig,  9),  the  formula  will  be, 
p  w  / 

D  —    Q     NX    *^      ^     * 


236  STRENGTH    OF    MATERIALS. 

When  a  beam  is  laid  in  a  horizontal  position,  fixed  at  both 
ends  and  loaded  in  the  middle  between  fastenings  (see  Fig.  10), 
the  formula  will  be, 

r>  Q   \s  -J    X    / 

f   —  o  A 

L  X  a 

When  a  beam  is  laid  in  a  horizontal  position,  fixed  at  both 
ends  and  the  load  evenly  distributed  over  its  whole  length  (see 
Fig.  11),  the  formula  will  be, 

P  —  12    X   ^  *  ^ 
L  X  a 

P  —  Breaking  load  in  pounds. 

.5"  =  Modulus  of  rupture,  which  is  72  times  the  weight,  in 
pounds,  which  will  break  a  beam  one  inch  square  and  one  foot 
long  when  fixed  in  a  horizontal  position,  as  shown  in  Fig.  (>, 
and  loaded  at  the  extreme  end,  and  which  may  be  taken  as 
follows  : 

Cast-Iron,  36,000. 

Wrought  Iron,  50,000. 

Spruce  and  Pine,  9,000. 

Pitch  Pine,  10,000. 

These  are  the  nearest  values,  in  round  numbers,  of  72  times 
the  average  value  of  the  constant  given  in  Table  No.  30. 

For  the  safe  load,  ^  may  be  taken  as  follows  : 
For  timber,  1,000  to  1,200  pounds. 
For  cast-iron,  3,000  to  5,000  pounds. 
For  wrought  iron,  10,000  to  12,000  pounds. 
For  steel,  12,000  to  20,000  pounds. 
L  =  Length  of  beam  in  inches. 
a  =•  The  distance  in  inches  from  the  neutral  surface  of 

the  section  to  the  most  strained  fiber. 
/  =  Rectangular  moment  of  inertia. 

The  tables  on  pages  237  and  238  give  the  moment  of  inertia 
about  the  neutral  axis  X  Y,  and  the  distance  a,  for  a  few  of  the 
most  common  sections : 

(For  explanation  of  moment  of  inertia  and  center  of  gravity 
see  page  293). 

These  formulas  have  the  great  advantage  of  being  theoretic- 
ally correct  for  beams  of  any  shape  of  cross-section,  made  from 
any  material,  providing  the  load  is  within  the  elastic  limit  of 
the  beam,  and  a  correct  constant  is  used  for  6"  and  the  correct 
value  obtained  for  the  moment  of  inertia. 


STRENGTH    OF    MATERIALS. 


237 


-H — « 


-Y    I 


2 

/ H 

a  -   6 


(*      B     * 


j 


12 


H 


-Vf 


r_H*x  B 
~12~ 

a=*L 
2 


a  6 


T 

9 

-aU 


36 


ai 
a 


12 


24 


« — d 


64 


a  = 


—  =  0.0982  Di  D* 


H 


f^(D*  —  d*)^ 

64 
Z> 


32 


32  D 


STRENGTH    OF    MATERIALS. 


"T! 


B  H*  —  b 


-Yr          X- 


12 


-Y.. 


tfB 

a  = 


T" 
f 


<p  j.     <?••*& 

ii  j 


l 


«i  = 


<2!   — 


B  H*  — 


2  (B  H  —  b  h} 
B  //2—  £,fc2 


,      (B  H'1  —  b  h  2)2  —  4  B  H  b  Jt  (H—hf 
12  (B  H  —  b  h) 

I  _  (B  H^  —  bh^  —  ^B  Hbh  (H  —  h}2 


/  _  (B  H*  —  b  h*f  —  4B  Hbh  (H  —  h)* 


STRENGTH    OF    MATERIALS. 


239 


Beams  of  symmetrical  section,  as  square,  round,  elliptical, 
or  H  section,  may  be  calculated  on  theoretically  correct  prin- 
ciples in  a  simpler  way,  obviating  the  use  of  the  moment  of 
inertia  and  the  modulus  of  rupture,  as  explained  below. 


P  — 


For  a  beam  fixed  at  one  end  and  loaded  at  the  other, 

ff-^ 


r,... 


CX  B 


L  = 


When  beam  is  square, 
Side  = 


B  = 


CX//2 


r  x  L 


When  beam  is  round, 

-v; 


Diameter 


X  L 


X  0.589 


P  =  Breaking  load  in  pounds. 

H  —  Thickness  or  height  of  beam  in  inches. 

B  —  Width  of  beam  in  inches. 

L  —  Length  of  beam  in  feet. 

C—  Constant  which  is  obtained  from  experiments,  and  is 
the  weight  in  pounds  which  will  break  a  beam  1  foot  long  and 
1  inch  square  fixed  at  one  end  and  loaded  at  the  other.  Con- 
stant C  is  given  in  Table  No.  30. 

A  rectangular  beam  fixed  at  one  end 
and  loaded  evenly  throughout  its  whole 
length  will  carry  twice  the  load  of  a  beam 
fixed  at  one  end  and  loaded  at  the  other; 
therefore, 

2  X  C  X  H*  X  B 


FIG.  7. 


P  =- 


L 


For  a  rectangular  beam  supported  under  both  ends  and 
loaded  at  the  center,  FIG.  B. 


P  = 


4  X  CX 


X  B 


A  rectangular  beam  supported  under  both  ends  and  loaded 
evenly  throughout  its  whole  length  will  carry  twice  the  load  of 


240 


STRENGTH    OF    MATERIALS. 


a  beam  supported  under  both  ends  and  loaded  at  the  center; 
therefore,  FIG.  9. 


P  = 


8  X  C  X 


X  B 


J3L 


For  a  beam  fixed  at  both  ends 
and  loaded  at  the  center, 


P  = 


X  B 


For  a  beam  fixed  at  both  ends 
and  the  load  distributed  evenly 
throughout  its  whole  length, 


P  = 


12  X  C  X  H2  X  B 


Each  letter  in  these  formulas  has  the  same  meaning  as  in 
formula  for  Fig.  6,  page  239,  and  each  formula  may  be  transposed 
the  same  as  that  formula.  The  most  convenient  way  is,  in  each 
case,  to  multiply  the  numerical  value  of  Cfrom  Table  No.  30,  by 
its  proper  coefficient  according  to  mode  of  loading,  before  it  is 
inserted  in  the  formula. 


NOTE. — A  square  beam  laid 
more  transverse  strength  than 
this  *  position. 


in  this  •  position  has  40  % 
the    same     beam     laid     in 


TABLE  No.  30.  —  Constant  C. 

Giving  the  weight  in  pounds  which  will  break  a  beam  one 
foot  long  and  one  inch  square  which  is  fastened  at  one  end,  in  a 
horizontal  position,  and  loaded  at  the  other  end. 


Material. 

Very  Good. 

Medium. 

Poor. 

Wrought  iron,* 

750 

600 

500 

Cast-iron, 

650 

500 

400 

Spruce  and  Pine, 
Pitch  pine, 

160 
225 

125 
150 

90 
100 

Granite, 

25 

*  A  wrought  iron  beam  or  bar  will  not  actually  break  under  these  conditions, 
but,  as-  it  will  bend  so  much  that  it  becomes  useless,  it  is  considered  to  be  equivalent 
to  the  breaking  point. 


STRENGTH    OF   MATERIALS. 


241 


The  following  formulas  will  apply  to  the  strength  of  beams 
of  the  shape  shown  in  the  adjacent  sectional  cuts.  These  for- 
mulas pertain  only  to  the  ultimate  breaking  strength  of  beams, 
and  have  nothing  to  do  with  deflection,  which  follows  entirely 
different  laws. 


SOLID   SQUARE   BEAMS. 


p  = 


CXH* 

FIG.  12 

r        PXL 

L 
CXH* 

H* 

3 
U              I    P   V    / 

X 

jjH 

HOLLOW  SQUARE  BEAMS. 


„ 


CX  (H*- 


FIG.  13. 


LXH 


SOLID   RECTANGULAR   BEAMS. 


__    CX  B  X  H*     FIG.  14 


L  — 


CX  B  X 


B  = 


s*i  


B   " 


PXL 


PXL 
B  x  H* 


P  — 


x  B 

HOLLOW  RECTANGULAR  BEAMS. 

B  X  H*  — 


HXL 


L—    CX(BXH^  —b  X  h"^ 
P  X  H 

SOLID   ROUND   BEAMS. 

p  =   0.589  CX  D*        FIG.  ie. 
0.589  C  X  D* 


FIG.  is. 


0.589  C 


242 


FIG.  17. 


FIG.  18. 


STRENGTH    OF    MATERIALS. 
HOLLOW   ROUND   BEAMS. 

p  _  0.589  CX(D*  —  d*) 
L  X  D 

SOLID  ELLIPTICAL   OR   OVAL  BEAMS. 


p_  0.589  C  X  Pi  X  D2 


HOLLOW  ELLIPTICAL  OR  OVAL  BEAMS. 


FIG.  19. 


P— 


0.589 


X  D*  —  </i  X 


LD 


FIG.  20. 


I    BEAMS. 

As  a  general  rule,  wrought  iron  I  beams 
should  always  be  selected  of  such  size  that  their 
depth  is  not  less  than  one-twenty-fourth  of  the 
span;  and  their  strength  may  be  calculated  by  the 
formula : 

P=  C  x  (B  x  H*  ~^b  x  h^ 
L  X  H 


In  the  preceding  formulas: 

P  =  Breaking  load  when  beam  is  fastened  at  one  end  and 
loaded  at  the  other. 

L  =  Length  of  beam  in  feet. 

C  =  Constant,  and  is  the  load  in  pounds  which  will  break  a 
bar  one  inch  square  and  one  foot  long  when  fastened  at  one 
end  and  loaded  at  the  other,  and  may  be  obtained  from 
Table  No.  30. 

These  formulas  give  the  breaking  load  when  the  beam  is 
fastened  at  one  end  and  loaded  at  the  other,  but  for  other 
fastenings  and  loads  C  must  be  multiplied  by  either  2,  4,  6,  8, 
or  12,  depending  upon  conditions.  (See  pages  239  and  240). 


STRENGTH    OF    MATERIALS.  243 

To  Find  the  Transverse  Strength  of  Beams  when  Their 

Section  is  Not  Uniform  Throughout  the 

Whole  Length. 

EXAMPLE. 

A  beam  made  of  wrought  iron  is  fastened  at  one  end  and 
loaded  at  the  other  (as  shown  in  Fig.  21).      The  largest  part 
is  5  inches   in   diameter  and  the  smallest  part  is  4  inches  in 
diameter.     Where   will  it  break? 
and  what  is  the  breaking  load  ?  FIG  21 

NOTE. — Naturally,  the  beam 
will  break  at  either  A  or  ^; 
therefore,  calculate  first  the  break- 
ing load  of  a  round  beam  of 
wrought  iron  4^  feet  long  and  5 
inches  in  diameter,  next  a  beam  3 
3  feet  long  (the  distance  from  P  to 
B}  .and  4  inches  in  diameter. 

Solving  for  strength  at  A  : 
p  _  D*  0.6  C 

53  X  0.6  X  600 


P  =  10,000  pounds. 
Solving  for  strength  at  B  : 
p  _  48  X  0.6  X  600 

3 
p  _  64  X  360 

3 
P  —  7680  pounds. 

Thus,  the  weakest  point  of  the  beam  is  at  B,  where  its  cal- 
culated breaking  load  is  only  7860  pounds,  while  the  calculated 
breaking  load  at  A  is  10,000  pounds. 

When  a  beam  is  not  of  uniform  section  throughout  its 
whole  length  and  is  supported  under  both  ends  and  loaded 
somewhere  between  the  supports,  calculate  first  the  reaction 
on  each  support ;  then  consider  the  beam  as  if  it  was  fastened 
by  the  load,  and  consider  the  reaction  at  each  support  as  a  load 
at  the  free  end  of  a  beam  of  length  and  section  equal  to  the 
length  and  section  between  its  load  and  support. 

EXAMPLE. 

The  largest  diameter  of  a  round  cast-iron  shaft  is  3  inches 
and  the  smallest  diameter  is  2  inches.  The  length,  mode  of 


2 44  STRENGTH  OF   MATERIALS* 

loading  aad  support  is   as  shown  in  Fig.  22.    Where  will  it 
break  ?  and  what  is  the  breaking  load  ? 


Solution  : 

The  reaction  at  y  will  be  Y*  of  the  load  and  the  reaction 
at  x  will  be  ft  of  the  load.  The  beam  will  evidently  break 
either  at  a,  b  or  n.  (Find  constant  C  in  Table  No.  30  and 
multiply  by  0.6,  because  the  beam  is  round). 

Solving  for  strength  at  n  : 
y  p  _  23  X  500  X  0.6 


P  — 


3 
X  300  X  8 


3X3 
P  —  2133^  pounds. 

Solving  for  strength  at  b  : 
„  _  33  X  500  X  0.6 


P  — 


5 

27  X  300  X  8 


5X3 
P  —  4320  pounds. 

Solving  for  strength  at  a  : 

s,    p  _  23  X  500  X  0.6 

/s  — 3— 

p  _  8  X  300  X  8 

5X2 
P  =  1920  pounds. 

Thus,  the  weakest  place  in  the  beam  is  at  a,  where  it  will 
break  when  loaded  at  b  with  1920  pounds.  If  the  load  is  moved 
nearer  ;z,  it  will  at  a  certain  point  exert  the  same  breaking  stress 
on  both  a  and  n. 

Regular  beams  of  this  kind  are  seldom  dealt  with,  but  shafts 
or  spindles  of  similar  shape  and  loaded  in  a  similar  manner  are 
frequently  used,  and  their  strength  and  stiffness  may  be  calcu- 
lated and  their  weakest  spot  ascertained  by  this  way  of  reason- 
ing, which  applies  as  well  to  hollow  as  to  solid  shafts  and 
spindles  made  from  wrought  iron,  steel  or  cast-iron. 


STRENGTH    OF    MATERIALS.  245 

Beams  fastened  at  one  end  and  loaded  at  the  other  may  be 
reduced  in  size  toward  the  loaded  end  and  still  have  the  same 
strength.     Suppose  the  beam  to  be  fastened  in  the  wall  at  X 
(Fig.  23)  and  loaded  at  the  other  end  with  a  given  load,  this  load 
will  then  have  the  greatest  breaking  effect  upon 
the  beam  at  X;  at  half  way  between  X  and  d 
the  load  has  only  half  the  breaking  effect,  at  c 
only  one-quarter  the  effect.   Therefore,  the  beam      d  c    b 
may  be  tapered  off  toward  b  in  such  proportion 
that  the  square   of  the  height   a  is  equal  to 
three-quarters  the  square  of  the  height  at  X. 
The  square  of  the  thickness  at  b  is  one-half  the 
square  of  the  thickness  at  X,  and  the  square  of  the  thickness  at 
c  is  one-quarter  the  square  of  the  height  at  X. 

EXAMPLE. 

An  iron  bracket  is  four  feet  long,  projecting  from  a  wall 
(as  Fig.  23).  It  is  strong  enough  when  24  inches  high  at  X. 
How  high  will  it  have  to  be  at  a,  b  and  £? 

Solution : 

X  =  242  —  576  Height  at  X  =  24" 

a  =  \/  %  X  576  =  \/432  Height  at  a  =  20.78" 
b  =  V  )4  X  576  =  \/288  Height  at  b  =  16.97" 
c  —  V  %  X  576  =  \/144  Height  at  c  —  12" 

The  curved  boundary  line  of  such  a  beam  is  a  parabolic 
curve,  because  the  property  of  a  parabola  is  that  the  square  of 
the  length  of  any  one  of  the  vertical  lines  (ordinates)  is  in  pro- 
portion as  their  distance  from  the  extreme  point  d.  By  this 
construction  one-third  of  the  material  may  be 
saved  and  the  same  strength  be  maintained.  F|G-  24- 

If  the  load  is  distributed  along  the 
whole  length  of  the  bracket  instead  of  at  its 
extreme  end,  it  would  have  the  form  shown  in 
Fig.  24. 

Square  and  Rectangular  Wooden  Beams. 

The  strength  increases  directly  as  the  width  and  as  the 
square  of  the  thickness.  The  strength  decreases  in  the  same 
proportion  as  the  length  of  the  span  increases. 

EXAMPLE  1. 

Find  the  ultimate  breaking  load  in  pounds  of  a  spruce 
beam  6  inches  square  and  8  feet  long,  when  supported  under 
both  ends  and  loaded  at  the  center. 


246  STRENGTH    OF   MATERIALS. 

Solution  : 

v_  4  CX  H* 

L 
^,^4X125X6X6X6 

~8~ 
P  =  13,500  pounds. 

NOTE.—  C  is  obtained  from  Table  No.  30,  and  is  multiplied 
by  4  because  the  beam  is  supported  under  both  ends  and  loaded 
at  the  center.  The  beam  is  square;  therefore  the  cube  of  the 
thickness  is  equal  to  the  square  of  the  thickness  multiplied  by 
the  width.  Consequently,  for  a  square  beam  (thickness)3  or 
(width)3  or  square  of  thickness  multiplied  by  width  is  the  same 
thing. 

EXAMPLE  2. 

Find  the  load  which  will  break  a  spruce  beam  8  inches 
thick,  4>£  inches  wide,  and  8  feet  long,  supported  under  both 
ends  and  loaded  at  the  center. 

Solution  : 

J,_±CXBXH* 


_4  X  125  X  4^  X  8  X  8 

8 

P  —  18,000  pounds. 
EXAMPLE  3. 

Find  the  load  which  will  break   the  beam  mentioned   in 
Example  2,  if  beam  is  laid  flatwise. 

_  4  X  125  X  8  X  ±y2  X  4^ 


P  —  10,125  pounds. 

In  the  first  example  the  beam  is  square,  6"  X  6"  =  36 
square  inches,  and  its  calculated  breaking  load  is  13,500  pounds. 
In  the  second  example  the  beam  is  rectangular,  8"  X  4>£"  '  = 
36  square  inches,  and  laid  edgewise  its  figured  breaking  load 
is  18,000  pounds.  In  the  third  example  the  same  beam  is  laid 
flatwise,  and  its  breaking  load  is  only  10,125  pounds.  Thus,  by 
making  a  beam  deep  it  is  possible  to  secure  great  strength  with 
only  a  small  quantity  of  material,  but  the  limit  is  soon  reached 
where  it  will  not  be  practical  to  increase  the  depth  at  the  ex- 
pense of  the  width,  because  the  beam  will  deflect  sidewise  and 
twist  and  break  if  it  is  not  prevented  by  suitable  means.  The 


STRENGTH    OF   MATERIALS.  247 

strongest  beam  which  can  be  cut  from  a  round  log  is  one  hav- 
ing the  thickness  1%  times  the  width.  The  stiff est  beam  cut 
from  a  round  log  has  its  thickness  lT7o  times  its  width.  The 
best  beam  for  most  practical  purposes  which  can  be  cut  from  a 
round  log  has  its  thickness  ll/2  times  its  width;  for  instance, 
4  X  6,  or  6  X  9,  or  8  X  12,  etc.  The  largest  side  in  a  beam 
having  its  thickness  \l/2  times  its  width  which  can  be  cut  from 
a  round  log  is  found  by  multiplying  the  diameter  by  0.832.  The 
diameter  required  in  a  round  log  to  be  large  enough  for  such  a 
beam  is  found  by  multiplying  the  largest  side  of  the  beam  by 
L.2;  for  instance,  the  diameter  of  a  round  log  to  cut  6"  X  9"  will 
be  9"  X  1.2  =  10.8  inches,  or  the  diameter  of  a  round  log  re- 
quired to  cut  8"  X  12"  will  be  12"  X  1.2  —  14.4  inches,  etc. 


To  Calculate  the  Size  of  Beam  to  Carry  a  Given  Load. 

Most  frequently  the  load  and  the  length  of  span  are  known 
and  the  required  size  of  beam  is  to  be  calculated.  For  a  rect- 
angular beam  there  would  then  be  two  unknown  quantities, 
the  width  and  the  thickness,  but  if  it  is  decided  to  use  a  beam 
having  its  thickness  1 l/2  times  its  width,  the  thickness  may  be 
expressed  in  terms  of  the  width. 

H  —  Thickness. 
B  —  Width. 


Use  formula  for  rectangular  beams,  page  239,  and  it  will 
read, 

„ C  X  (l/^  -B)2  X  B 

L 
This  will  reduce  to, 

C  X  2X  X  B9 

1       


This  will  transpose  to, 


C  X  2# 
EXAMPLE. 

Find  width  and  thickness  of  a  spruce  beam  10  feet  long, 
when  fastened  at  one  end  and  required  to  carry,  with  8  as  factor 
of  safety,  a  load  of  1800  pounds  at  the  other  end,  the  thickness 
to  be  1  l/2  times  the  width. 

When  the  beam  is  to  carry  1800  pounds,  with  8  as  a  factor 
of  safety,  its  breaking  load  is  8  X  1800  =  14,400  pounds. 


248  STRENGTH    OF   MATERIALS. 

Solution  : 


B— 

B  —  S  inches  in  width. 

H—\yz  X  8"  =  12  inches  in  thickness. 

The  weight  of  the  beam  itself  is  not  considered  in  this 
problem. 

To  Find  the  Size  of  a  Beam  to  Carry  a  Given  Load  When 
Also  the  Weight  of  the  Beam  is  to  be  Considered. 

RULE. 

Calculate  first  the  size  of  beam  required  to  carry  the  load? 
then  figure  what  such  a  beam  will  weigh  and  add  half  of  this 
weight  to  the  load,  if  the  beam  is  fastened  at  one  end  and  loaded 
at  the  other,  or  supported  under  both  ends  and  loaded  at  the 
center,  but  add  the  whole  weight  of  the  beam  to  the  weight  of 
the  load  if  the  load  is  distributed  along  the  whole  length  of  the 
beam.  Then  figure  the  size  of  the  required  beam  for  this  new 
load. 

EXAMPLE. 

Find  width  and  thickness  of  a  pitch  pine  beam  to  carry 
2000  pounds,  with  8  as  factor  of  safety,  and  a  span  of  27  feet. 
The  beam  is  supported  under  both  ends  and  loaded  at  the  center  ; 
its  own  weight  is  also  to  be  taken  into  consideration. 

Solution  : 

Find  the  constant  for  pitch  pine  in  Table  No.  30  to  be  150, 
and  find  the  weight  of  pitch  pine  in  Table  No.  10  to  be  50  pounds 
per  cubic  foot.  When  the  beam  is  supported  under  both  ends 
and  loaded  at  the  center  it  is  four  times  as  strong  as  if  fastened 
at  one  end  and  loaded  at  the  other;  therefore,  constant  150  is 
multiplied  by  4.  The  load,  2000  pounds,  multiplied  by  8  as  a 
factor  of  safety,  gives  10,000  pounds  as  breaking  load  of  the 
beam. 


„      ^ ,      16000  X  27 

j£>  = 


2X   X  150  X  4 
3  

B  =  6.84 "  =  width,  and  1#  X  6.84"  =  10.26"  =  thickness. 

The  area  is  6.84  X  10.26  =  70  square  inches  ;  the  weight  per 

foot  is  70  times  50  divided  by  144,  which  equals  24.3  pounds,  say  25 

pounds.     The  weight  of  the  beam  is  25  X  27  =  675  pounds.     This 


STRENGTH    OF   MATERIALS.  249 

weight  is  distributed  along  the  whole  beam  and,  therefore,  it  does 
not  have  any  more  effect  than  if  half  of  it,  or  337  J^  pounds, 
was  placed  at  the  center,  but  as  the  beam  is  to  be  calculated  with 
8  as  factor  of  safety,  the  weight  allowed  for  the  beam  must  be 
337^2  X  8  =  2700  pounds.  Thus,  adding  this  weight  to  16,000 
pounds  gives  18,700  pounds  ;  this  new  weight  is  used  for  calcu- 
ing  the  size  of  the  required  beam. 


J5  = 


374 

B  =  7.2  inches  =  width,  and  IY2  X  7.2ff  =  10.8  inches, 
thickness. 

This,  of  course  is  also  a  little  too  small,  as  only  the  weight 
of  a  beam  6.84  inches  by  10.25  inches  is  taken  into  account,  but 
if  more  exactness  should  be  required  the  weight  of  this  new 
beam  may  be  calculated  and  the  whole  figured  over  again, 
and  the  result  will  be  closer.  This  operation  may  be  repeated 
as  many  times  as  is  wished,  and  the  result  will  each  time  be 
closer  and  closer,  but  never  exact;  but  for  all  practical  purposes 
one  calculation,  as  shown  in  this  example,  is  sufficient. 

EXAMPLE  2. 

Find  width  and  thickness  of  a  spruce  beam  to  carry  4200 
pounds  distributed  along  its  whole  length.  The  span  is  24  feet ; 
use  10  as  factor  of  safety,  and  also  allow  for  the  weight  of  beam. 
The  thickness  of  the  beam  is  to  be  1%  times  its  width. 

Solution : 

3 

/  4200  X  24  X  10 
>  2X   X  125  X  8 


B  =  7.65  inches,  and  H—  11.48  inches. 

iir  •  u,.    r  u  7-65  X  11.48  X  24  X  32 

Weight  of  beam  =  —         — — —  —  =  468  pounds. 

144 

Adding  ten  times  the  weight  of  the  beam  to  ten  times  the 
weight  to  be  supported,  gives  46,680  pounds. 

3 

„  /     46680  X  24 


2X  X  125  X  8 


3 

—  V  497.9 


/?  =  7.93  inches,  and  H  —  \%  B=  11.9  inches,  or  prac- 
tically, a  beam  8  inches  by  12  inches  is  required. 


250  STRENGTH   OF   MATERIALS. 

Crushing  and  Shearing  Load  of  Beams  Crosswise  on  the 

Fiber. 

Too  much  crushing  load  must  not  be  allowed  at  the  ends  of 
the  beams  where  they  rest  on  their  supports,  as  all  kinds  of 
wood  has  comparatively  low  crushing  strength  when  the  load  is 
acting  crosswise  on  the  fiber. 

Approximately,  the  average  ultimate  crushing  strength  of 
wood,  crosswise  of  the  fiber,  is  as  follows  : — 

White  oak,  2000  pounds  per  square  inch. 

Pitch  pine,  1400  pounds  per  square  inch. 

Chestnut,  900  pounds  per  square  inch. 

Spruce  and  pine,  500  to  1000  pounds  per  square  inch. 

Hemlock,  500  to  800  pounds  per  square  inch. 

The  safe  load  may  be  from  one-tenth  to  one-fifth  of  the 
ultimate  crushing  load.  When  the  wood  is  green  or  water- 
soaked,  its  crushing  strength  is  less  than  is  given  above. 

EXAMPLE. 

How  much  bearing  surface  must  be  allowed  under  each 
end  of  the  beam  mentioned  in  Example  2,  providing  it  also 
has  10  as  a  factor  of  safety  ?  The  crushing  strength  of  spruce 
crosswise  on  the  fiber  is  500  pounds,  and  using  10  as  factor  of 
safety,  the  load  allowed  per  square  inch  must  be  only  50  pounds. 
The  beam  is  8  inches  wide,  and  half  of  4685  pounds  is  sup- 
ported at  each  end ;  thus  the  length  of  bearing  required  under 

2:}42 
each  end  will  be  _0  x  g  =  5.85  inches.     Thus,  the  least  bearing 

allowable  should  be  about  6  inches  long. 

When  beams  are  heavily  loaded  and  resting  on  posts,  or 
have  supports  of  small  area,  either  hardwood  slabs  or  cast-iron 
plates  should  be  placed  under  their  ends,  in  order  to  obtain 
sufficient  bearing  surface  for  the  soft  wood. 

The  same  care  must  be  exercised  when  a  beam  is  loaded  at 
one  point;  the  bearing  surface  under  the  load  should  at  least 
be  as  long  as  the  bearing  surface  of  both  ends  added  together. 

Short  beams  are  liable  to  break  from  shearing  at  the  point 
of  support,  especially  when  loaded  throughout  their  whole  length 
to  the  limit  of  their  transverse  strength. 

The  ultimate  shearing  strength  for  spruce,  crosswise  of  the 
fiber,  is  3000  pounds  per  square  inch  (see  page  273).  Safe  load 
may  be  300  pounds  per  square  inch. 

In  the  above  example  the  beam  is  8"  X  12"  —  96  square 
inches,  and  its  center  load  is  4085  pounds,  or  2342 ]/2  pounds  at 

2342 1/ 

each  end.  The  shearing  stress  is  -^p  =  24.6  pounds  per 
square  inch.  Hence,  the  factor  of  safety  against  shearing  is 
about  100,  and  there  is  not  the  least  danger  that  this  beam  will 
give  way  under  shearing;  but  such  is  not  always  the  result* 


STRENGTH    OF    MATERIALS.  25  1 

Round  Wooden  Beams. 

A  round  beam  has  0.589  times  the  strength  of  a  square 
beam  of  same  length  and  material,  when  the  diameter  is  equal 
to  the  side  of  the  square  beam.  The  area  of  a  square  beam 
compared  to  the  area  of  the  round  beam  is  as  0.7854  to  1  ;  there- 
fore it  might  seem  as  if  that  also  should  be  the  proportion  be- 
tween their  strength,  which  is  the  case  for  tensile,  crush- 
i  ig  and  shearing  strength,  but  not  for  transverse  strength 
or  for  deflection,  because  the  material  is  not  applied  to  such 
advantage  in  the  round  beam  as  it  is  in  the  square  one.  All 
preceding  formulas  for  transverse  strength  of  square  beams  may 
also  be  used  for  round  beams  if  only  constant  C  is  multiplied 
by  0.589,  or,  say,  0.6. 

Thus,  the  formula  for  a  round  beam  fastened  at  one  end 
and  loaded  at  the  other  will  be  : 

0.6  C  X  Z>3 


NOTE.  —  In  a  round  beam,  of  course,  it  will  be  Z>3  instead  of 
//"2  b  for  a  rectangular  one. 

EXAMPLE. 

Find  the  load  in  pounds  which  will  break  a  spruce  beam  12 
feet  long  and  0  inches  in  diameter  when  supported  under  both 
ends  and  loaded  at  the  center.  (Find  constant  C  in  Table  No.  30.) 

Solution  : 

4  X  0.6  C  X  Z>3 


L 
4  X  0.6  X  125  X  6  X  6  X  6 

12 
P  =  5400  pounds. 

To  Calculate  the  Size  of  Round  Beams  to  Carry  a  Given 
Load  When  Span  is  Known. 

Where  the  load  and  span  are  known,  the  diameter  of  the 
beam  is  calculated,  when  fastened  at  one  end  and  loaded  at  the 
other,  by  the  formula  : 


_     iP  XLX  factor  of  safety 
=  >~~ 


0.6  C 
RULE. 


Multiply  together  the  load  in  pounds,  factor  of  safety  and 
length  of  span  in  feet,  divide  this  product  by  six-teji<;hs  of  the 


25 


STRENGTH    OF    MATERIALS. 


constant  in  Table  No.  30,  and  the  cube  root  of  this  quotient  is 
the  diameter  of  the  beam. 

EXAMPLE. 

A  round  spruce  beam  is  fastened  into  a  wall,  and  is  to 
carry  1200  pounds  on  the  free  end  projecting  4  feet  from  the 
wall,  with  8  as  a  factor  of  safety,  the  weight  of  the  beam  not  to 
be  considered.  Find  diameter  of  beam. 

Solution : 


1200  X  4  X  8 
0.6  X  125 


38400 
75 


—  V  512 
D  =  8  inches  diameter. 


Load  Concentrated  at  Any  Point,  Not  at  the  Center  of 
a  Beam. 

If  a  beam  is  supported  at  both  ends  and  loaded  anywhere 
between  the  supports  but  not  at  the  center  (see  Fig.  25),  it  will 
carry  more  load  than  if  it  was  loaded  at  the  center.  With 
regard  to  breaking,  the  carrying  capacity  is  inversely  as  the 
square  of  half  the  beam  to  the  product  of  the  short  and  the  long 
ends  between  the  load  and  the  support.  For  instance,  a  beam 
10  feet  long  is  of  such  size  that  when  it  is  supported  under 
both  ends  and  loaded  at  the  center  it  will  carry  1400  pounds. 
How  many  pounds  will  the  same  beam  carry  if  loaded  3  feet 
from  one  end  and  7  feet  from  the  other  ? 

Solution : 

^      1400  X  52 


y  — 


7X3 

1400  X  25 

21 


FIG.  25. 


X=  1666%  pounds. 
If  weight  of  beam  is  also  in- 
cluded in  its  center-breaking-load, 
the  formula  will  be  : 


P  —  Breaking  load  (including 

weight  of  beam)  if  applied  at  the  center  in  pounds. 

F  =  Half  the  length  of  the  span. 


STRENGTH    OF  MATERIALS. 


253 


JF=  Weight  of  beam. 

P\  =  Breaking  load  applied  at  n. 

Load  at  n  X  distance  b 


Load  on  Pier  A  = 


Load  on  Pier  B  = 


span 

Load  at  n  X  distance  a 
span 


Beams  Loaded  at  Several  Places, 

FIG.  26. 

"i       *  -s 


1 


i 


10-ffa 


When  a  beam  is  loaded  at  several  places  the  equivalent 
center  load  and  the  load  on  each  support  may  be  calculated  as 
shown  in  the  following  example:  (See  Fig.  26). 

The  equivalent  center  load  for  a  =   4  X  2--  X  1000  =  555.6  Ibs. 

12  X  12 


The  equivalent  center  load  for  b  = 


10  X 


12  X  12 


X    800  =  777.8  Ibs. 


The  equivalent  center  load  for  c  =  1f  X    8  X    900  =  800  lbs. 

The  equivalent  center  load  for  d  —  19  X    5  X    300  =  197.9  lbs. 

12  X  12 

The  equivalent  center  load  for  loads  a,  b,  c  and  d  is  2331.3 
pounds. 

The  load  on  Pier  A  = 

(5  X  300)  +  (8  X  900)  +  (14  X  800)  +  (20  X  1000)  _  WQ2y  lbs 

24 

The  load  on  Pier  B  = 
(4  X  1000)  +  (10  X  800)  +  (16  X  900)  +  (19  X  300)  =  1337  y  lbs 

24 

NOTE. — The  sum  of  the  load  on  supports  A  and  B  is  always 
equal  to  the  sum  of  all  the  loads ;  therefore,  by  subtracting  the, 


254 


STRENGTH    OF    MATERIALS. 


calculated  load  on  B  from  the  total  load  the  load  on  A  is  ob- 
tained. By  subtracting  the  calculated  load  at  A  from  the  total 
load,  the  load  on  B  is  obtained. 

To  each  load  as  calculated  above  for  each  support  also  add 
half  the  weight  of  the  beam. 


To  Figure  Sizes   of 


FIG.  23. 


Beams  When  Placed  in  an  Inclined 
Position. 

Figure  all  calculations  concerning 
the  transverse  strength  from  the  dis- 
tance S,  and  leave  the  length  L  out  of 
consideration.  If  the  distance  S  cannot 
be  obtained  by  measurement  it  may 
be  found  by  multiplying  L  by  cosine  of 
angle  a. 


DEFLECTION    IN    BEAMS    WHEN    LOADED 
TRANSVERSELY. 

Experiments  and  theory  both  prove  that  if  the  span  is 
increased  and  the  width  of  the  beam  increased  in  the  same  pro- 
portion the  transverse  strength  of  the  beam  is  unchanged  ;  but 
such  is  not  the  case  with  its  stiffness.  If  a  beam  is  to  have  the 
same  stiffness  its  depth  must  be  increased  in  the  same  ratio  as 
the  span,  providing  the  width  is  unchanged.  Within  the 
elastic  limit  of  the  beam  the  deflection  is  directly  proportional 
to  the  load;  that  is,  half  the  load  produces  half  the  deflection, 
out  doubling  the  load  will  double  the  deflection. 

Deflection  is  proportional  to  the  cube  of  the  span  ;  that  is, 
with  twice  the  length  of  span  the  same  load  will,  when  the  other 
dimensions  of  the  beam  are  unchanged,  produce  eight  times  as 
much  deflection. 

Deflection  is  inversely  as  the  cube  of  the  depth  (thickness) 
of  the  beam.  For  instance,  if  the  depth  of  a  beam  is  doubled 
but  the  length  of  span  and  the  width  of  beam  is  unchanged,  the 
same  load  will  produce  only  one-eighth  as  much  deflection. 
Deflection  is  inversely  as  the  width  of  the  beam ;  for  instance, 
when  a  beam  is  twice  as  wide  as  another  beam  of  the  same 
material  but  all  the  other  dimensions  are  unchanged,  the  same 
load  will  produce  only  half  as  much  deflection. 

The  deflection  in  a  beam  caused  by  various  modes  of  load- 
ing is  calculated  by  the  following  formulas  : — 

For  beams  laid  in  a  horizontal  position  and  loaded  trans- 
versely, fastened  at  one  end  and  loaded  at  the  other :  (See  Fig.  6). 


3X^X7 


STRENGTH    OF    MATERIALS.  255 

For  beams  laid  in  a  horizontal  position,  fastened  at  one  end 
and  loaded  thoroughout  the  whole  length  :    (See  Fig.  7.) 

P  X  Z3 


C* " 


8  X  E  X  S 


For  beams  laid  in  a  horizontal  position,  supported  under  both 
ends  and  loaded  at  the  center:     (See  Fig.  8). 

S  — 


48  X  E  X  I 

This  formula  may  be  transposed  and  used  to  calculate 
modulus  of  elasticity  from  the  results  obtained  when  specimens 
are  tested  for  transverse  stiffness.  Deflection  should  be  care- 
fully measured  but  the  specimen  must  not  be  bent  beyond  its 
elastic  limit  ;  the  modulus  of  elasticity  is  calculated  by  the  trans- 
posed formula  : 


48  X  S  X  / 

For  a  square  specimen  /  is  (side  of  beam)4  divided  by  12. 
(See  moment  of  inertia,  page  237). 

(Also  see  rule  for  calculating  modulus  of  elasticity, 
page  265). 

For  beams  laid  in  a  horizontal  position,  supported  under  both 
ends  and  loaded  uniformly  throughout  their  whole  length  :  (See 
Fig.  D). 

S=    5  X  P  X  L* 
384  XE  X  r 

For  beams  laid  in  a  horizontal  position,  fixed  at  both  ends, 
and  loaded  at  the  center  :  (See  Fig.  10). 

S= 


192  XEXf 

For  beams  laid  in  a  horizontal  position,  fixed  at  both  ends  and 
loaded  uniformly  throughout  their  whole  length  :     (See  Fig.  11). 


384  X  E  X  / 

In  these  formulas  the  definitions  of  the  letters  are  : 
S  =  Deflection  in  inches. 
P  =  Load  in  pounds. 
L  =  Length  of  span  in  inches. 

E  •=•  Modulus  of  elasticity  in  pounds  per  square  inch. 
/  =  Rectangular  moment  of  inertia.    (See  pages  237-238). 

These  formulas  are   applicable    to  any  shape  of  section 
or  material,  when  the  load  is  within  the  elastic  limit. 


256  STRENGTH    OF   MATERIALS. 

For  beams  of  symmetrical  section  it  is  more  convenient  to 
use  the  following  equally  correct  but  more  practical  formulas,  by 
which  the  deflection  is  calculated  directly  from  the  size  of  the 
beam  by  simply  using  a  constant  obtained  by  experiment  and 
reduced  by  calculation  to  a  unit  beam  one  foot  long  and  one 
inch  square,  thus  avoiding  both  the  use  of  the  modulus  of  elas- 
ticity and  the  moment  of  inertia. 

When  beams  are  supported  under  both  ends  and  loaded  at 
the  center,  and  the  weight  of  the  beam  itself  is  not  considered, 
the  following  formulas  may  be  used  for  solid  rectangular  beams 
laid  in  a  horizontal  position  : 

?  _  L8  X  P  X  c  3  _ 

H*XB  L  =.  J//3  X  B  XS 

^  ' 


X  B  X  c  S  X  H*  X  B 


S  X  B  L*X  P 

B  _  Z,3  X  P  X  c  p  _  H*  X.  B  X  S 

SXH*  L*Xc 

S  =  Deflection  in  inches. 

ff=  Thickness  of  beam  in  inches. 

B  •=.  Width  of  beam  in  inches. 

L  =  Length  of  beam  in  feet. 

P  =  Load  in  pounds. 

c  —  Constant  obtained  by  experiment,  and  is  the  deflec- 
tion, in  fractions  of  an  inch,  which  a  beam  one  foot  long  and 
one  inch  square  will  have  if  supported  under  both  ends  and 
loaded  at  the  center;  the  average  value  for  this  constant  is 
given  in  Table  No.  31. 

For  any  other  mode  of  loading,  see  rules  and  explanations 
on  page  261. 

In  previous  formulas  and  rules,  the  weight  of  the  beam 
itself  was  not  considered.  The  deflection  in  a  beam  caused  by 
its  own  weight  when  it  is  of  rectangular  shape  and  uniform 
size,  and  laid  in  a  horizontal  position,  is  obtained  by  the 
formula, 

Z3  X  #  W  X  c 
H*XB 

When  both  the  weight  and  the  load  are  to  be  considered, 
the  deflection  in  a  solid  rectangular  beam  laid  in  a  horizontal 
position,  supported  under  both  ends  and  loaded  at  the  center,  is 
calculated  by  the  formula, 


H*X  B 


STRENGTH    OF   MATERIALS. 


257 


6*  =  Deflection  in  inches. 

L  =  Length  of  span  in  feet. 

P  =  Load  in  pounds. 

W—  Weight  of  beam  in  pounds. 

c  —  Constant  obtained  by  experiments,  and  is  the  deflec- 
tion in  fractions  of  an  inch,  which  a  beam  one  foot  long  and 
one  inch  square  will  have  if  supported  under  both  ends  and 
loaded  at  the  center,  and  may  be  found  in  Table  No.  31. 

//=  Thickness  of  beam  in  inches. 

B  =  Width  of  beam  in  inches. 

RULE. 

To  the  load  add  five-eighths  of  the  weight  of  beam,  mul- 
tiply this  by  the  cube  of  the  length  of  the  span  in  feet,  and 
multiply  by  a  constant  from  Table  No.  31.  Divide  this  product 
by  the  product  of  the  cube  of  the  thickness  and  the  width  of 
the  beam ;  the  quotient  is  the  deflection  in  inches. 

The  deflection  in  a  beam  supported  under  both  ends  and 
loaded  evenly  throughout  is  five-eighths  of  that  of  a  beam 
supported  under  both  ends  and  loaded  at  the  center.  Therefore, 
in  the  following  formulas,  the  weight  of  the  beam  itself  is  multi- 
plied by  five-eighths  to  reduce  the  effect  of  the  weight  of  the 
beam  to  the  equivalent  of  a  load  placed  at  its  center. 

FOR  SOLID   SQUARE  BEAMS. 


FIG.  28 


FOR  SOLID  RECTANGULAR  BEAMS. 


B  H* 


-\ 

x  FIG.  29 

I 


B 


FOR  HOLLOW  SQUARE  BEAMS. 


FIG.  3O 


STRENGTH   OF   MATERIALS. 
FOR   HOLLOW   RECTANGULAR  BEAMS. 


FOR  I  BEAMS. 


FIG.  31 


—  2  b  h* 


•b- 


FIG.  32 


c^.g 


FOR  SOLID  ROUND  BEAMS. 

h  X  W\  Z3 


Z* 

FOR  HOLLOW  ROUND  BEAMS. 
1.7<:CP  +  &  W)  Z3 


FIG.  33 


FIG.  34 

FOR    SOLID   ELLIPTICAL   OR   OVAL  BEAMS. 
S  =  ± 


Dl  X 


FIG.  35 


FOR  HOLLOW  ELLIPTICAL  OR  OVAL  BEAMS. 


1.7 


STRENGTH   OF    MATERIALS. 


S  =  Deflection  in  inches. 

L  —  Length  of  span  in  feet. 

P  =  Load  in  pounds. 

W=-  Weight  of  beam  in  pounds. 

c  =  Constant  obtained  from  experiments,  or  may  be  ob- 
tained from  Table  No.  31. 

For  meaning  of  the  other  letters,  see  figure  opposite  each 
formula. 

A  round  beam  equal  in  diameter  to  the  side  of  a  square 
beam  will  deflect  1.698  times  as  much,  and  for  convenience, 
when  the  deflection  of  a  square  or  a  rectangular  beam,  whether 
solid  or  hollow,  is  known,  it  may  be  multiplied  by  1.7,  and  the 
product  is  the  deflection  of  a  corresponding  round,  oval,  or 
elliptical  beam  of  the  same  material  and  diameter  and  laid  in 
the  same  relative  position  and  loaded  in  the  same  manner  as 
the  calculated  beam.  It  is  well  to  remember  that  a  round  or 
elliptical  beam  weighs  a  little  less  than  a  square  or  rectangular 
one,  when  the  sides  and  diameters  are  equal,  and  the  deflection 
due  to  its  own  weight  is,  therefore,  a  little  less. 


TABLE  No.  3 1.— Constant  c, 

Giving  deflection  in  inches  per  pound  of  load  when  the  beam 
is  supported  under  both  ends  and  loaded  at  the  center. 


MATERIAL. 

Constant  c. 

MATERIAL. 

Constant  c. 

Cast  steel, 
Wrought  iron, 
Machinery  steel 
Cast-iron, 

0.0000143 
0.0000156 
O.C000156 
0.0000288 

Pitch  pine, 
Spruce, 
Pine, 

0.00024 
0.00035 
0.00033 

EXAMPLE. 

A  beam  6"  X  9"  of  pitch  pine,   10  feet  long,  supported 
under  both  ends,  is  to  be  loaded  at  the  center  with  one-tenth 
of  its  breaking  load.     Find  the  load  and  deflection. 
Solution  : 

_  92  X  6  X  4  X  150 
10  X  10 


Deflection  will  be, 
s=  10*  X  2916^0.00024     =   699.84  = 


260  STRENGTH    OF   MATERIALS. 

Therefore,  if  this  beam  had  been  curved  0.16  inch  upward, 
by  increasing  its  thickness  on  the  upper  side,  it  would  have  been 
straight  after  the  load  was  applied.* 

In  this  example  the  weight  of  the  beam  itself  is  not 
considered  either  in  figuring  the  strength  or  the  deflection, 
because  the  beam  is  comparatively  short  in  proportion  to  its 
width  and  thickness.  The  weight  of  the  beam  itself  will  only 
be  about  200  pounds,  and  this  will  be  of  no  account  in  propor- 
tion to  the  load  that  the  beam  will  carry,  with  10  as  a  factor  of 
safety.  The  weight  of  the  beam  will  increase  its  deflection  cnly 
0.006  inch.  In  such  a  beam  the  danger  is  probably  greater  from 
crushing  of  the  ends  at  the  supports,  if  it  has  not  enough  bear- 
ing surface.  In  long  beams  the  weight  of  the  beam  must  not 
be  neglected,  either  in  calculating  safe  load  or  in  calculating 
deflection. 

EXAMPLE  2. 

A  round  bar  of  wrought  iron  is  5  feet  long  and  3  inches  in 
diameter,  and  loaded  at  the  center  with  800  pounds.  How  much 
will  it  deflect  ?  A  round  bar  of  iron  3  inches  in  diameter  and 
5  feet  long  weighs  119  pounds.  (See  table  of  weights  of  iron, 
page  143.) 

Solution : 

e  _  53  X  (800  -f  #  X  119)  X  1.7  X  0.0000156 

~3*~ 
S  =  0.0359  inch. 

Thus,  such  a  shaft  loaded  with  800  pounds  will  deflect  T£-§7 
of  an  inch  in  the  length  of  5  feet,  or  60  inches.  If  the  deflec- 
tion must  not  exceed  ir^  of  the  span  (see  page  266),  then  the 
greatest  allowable  deflection  for  this  span  would  be  0.04  inch, 
and  the  calculated  deflection  is  within  this  limit. 

NOTE. — T3W  of  the  span  is  equal  to  a  deflection  of  0.008 
inch  per  foot  of  length. 

EXAMPLE  2. 

A  shaft  of  machinery  steel,  11  inches  in  diameter  and  6  feet 
between  bearings,  carries  in  the  center  a  12-ton  fly  wheel.  How 
much  deflection  will  the  weight  of  the  fly  wheel  cause? 

NOTE. — Such  shafts  are  usually  considered  as  a  beam  sup- 
ported under  both  ends.  (See  formula  for  deflection  in  solid 
round  beams,  page  258.) 

Solution : 

12  tons  =  24,000  pounds.  (Weight  of  shaft  is  not  taken 
into  consideration.) 

*  This  is  a  thing  frequently  done  in  practice. 


STRENGTH    OF    MATERIALS.  261 

X  1.7  c 


63  X  24000  X  1.7  X  0.0000156 
II4 

e_  216  X  24000  X  0.00002652 

14641 

S  =  0.00939  inches. 

Thus,  the  calculated  deflection  caused  by  the  fly  wheel  is  a 
little  less  than  Tfa  of  an  inch.  The  deflection  per  foot  of  span 
will  be  Q,-Qo^9_-3  9  which  equals  0.001565  inch. 

EXAMPLE  3. 

Calculate  the  deflection  of  shaft  mentioned  in  the  previous 
example,  when  both  the  weight  of  fly  wheel  and  the  weight  of 
shaft  are  to  be  considered. 

Solution : 

s  _  68  X  (24000  -f  H  X  1920)  X  1.7  X  0.0000156 

_  216  X  25200  X  0.00002652 
o  — 


14641 
.S"  =  0.00986    inch. 

Practically,  the  deflection  is  likely  to  be  a  little  less  than 
what  is  figured  in  the  two  previous  examples,  because  if  the 
hub  of  the  fly  wheel  fits  well  on  the  shaft,  it  will  stiffen  it  some. 
(It  is  a  g^ood  practice  to  make  such  shafts  a  little  larger  in 
diameter  in  the  place  where  the  hub  of  the  wheel  is  keyed  on  ; 
this  enlargement  will  then  compensate  for  what  the  shaft  is 
weakened  by  cutting  the  key-way.) 

The  weight  of  the  shaft  may  be  obtained  by  considering  a 
cubic  foot  of  machinery  steel  to  weigh  485  pounds,  and  a  shaft 
11  inches  in  diameter  will  then  weigh  320.1  pounds  per  foot  in 
length,  and  6  feet  will  weigh  1920  pounds.  Multiplying  this  by 
#  gives  1200  pounds,  to  be  added  to  the  weight  of  the  fly  wheel, 
which  gives  25,200  pounds.  The  weight  of  the  shaft  may  also 
be  found  in  the  table  of  weight  of  round  iron,  page  144. 

To  Calculate  Deflection  in  Beams  Under  Different  Modes 
of  Support  and  Load. 

Constant  c  in  Table  No.  31  is  the  deflection  in  fractions  of 
an  inch  per  pound  of  load  when  a  beam  one  foot  long  and  one 
inch  square  is  supported  under  both  ends  and  loaded  at  the 
center,  and  when  this  constant  for  any  given  material  is  known, 
the  deflection  for  beams  subjected  to  other  modes  of  fastening 
and  loads  may  be  calculated  thus  : 

For  beams  supported  under  both  ends  with  the  load  dis- 
tributed evenly  throughout  their  whole  length,  multiply  c  by  # . 


262  STRENGTH   OF   MATERIALS. 

For  beams  fixed  at  both  ends  and  loaded  at  the  center, 
multiply  c  by  #. 

For  beams  fixed  at  both  ends  with  the  load  distributed 
evenly  throughout  their  whole  length,  multiply  c  by  y%. 

For  beams  fixed  at  one  end  and  loaded  at  the  other,  mul- 
tiply c  by  16. 

For  beams  fixed  at  one  end  with  load  distributed  evenly 
throughout  their  whole  length,  multiply  c  by  6. 

EXAMPLE. 

A  square,  hollow  beam  of  cast-iron,  8  inches  outside  and  6 
inches  inside  diameter,  and  9-foot  span,  supported  under  both 
ends,  is  loaded  at  the  center  with  8000  pounds.  How  much 
will  it  deflect  ? 

Solution : 

Weight  of  beam  =  9  X  12  X  (82  —  62)  X  0.26  —  786  pounds. 


.S1  =  g3  X  (800°  +  ^  x  786)  x  0.0000288 

8*  — 6* 
y_729  X  8492  X  0.0000288 

4096  —  1296 
s_  178.291 

2800 
^=0.064  inches. 

EXAMPLE. 

How  much  would  this  same  beam  deflect  if  the  load  had 
been  distributed  evenly  throughout  its  whole  span? 
Solution  : 

S  =  L*  (P  +  W">  5/8  c 
s  _  93  X  8786  X  ^  X  0.0000288 
115.289 


2800 

.9  =  0.041  inch. 
EXAMPLE. 

A  round  cast-iron  beam  of  7  inches  outside  and  5  inches 
inside  diameter  is  4  feet  between  supports,  with  a  load  of  2000 
pounds  distributed  evenly  throughout  its  span.  How  much  will 
it  deflect,  the  weight  of  beam  itself  not  being  considered  in  the 
calculation  ? 


STRENGTH    OF   MATERIALS. 


263 


Solution : 

o      43  X  2000  X  0.0000288  X  1.7  X 


74  _  54 

S  =  256  X  200°  X  0.0000288  X  1.7  X  ft 

1776^ 
^  =  0.0085  inch. 

In  this  example,  1.7  is  used  as  a  multiplier  because  the 
beam  is  round,  and  ^  because  the  load  is  distributed  evenly 
throughout  the  length  of  the  span. 

EXAMPLE. 

A  fly  wheel  weighing  800  pounds  is  carried  on  the  free  end 
of  a  3-inch  shaft,  1  foot  from  the  bearing.  How  much  will  the 
shaft  deflect? 

This  is  the  same  as  a  round  beam  loaded  at  one  end  and 
fastened  at  the  other;  therefore,  constant  c  is  multiplied  by 
16  X  1.7. 

Solution : 

~      L8  P  1.7  c  X  16 


C*  


1  X  800  X  1.7  X  0.0000156  X  16 


S  =  0.0042  inch. 

Previous  calculations  for  breaking  load  and  also  for 
deflection  are  based  upon  a  dead  load  slowly  applied  and  not 
exposed  to  jar  and  vibrations.  If  the  load  is  applied  suddenly 
it  will  have  greater  effect  toward  breaking  the  beam  than  ff 
applied  slowly.  For  instance,  imagine  a  load  having  its  whole 
weight  hanging  on  a  rope,  like  Fig.  37,  just  touching  the  beam 
but  not  actually  resting  upon  it. 
If  that  rope  was  cut  off  suddenly 
this  load  would  produce  twice  as 
much  effect  toward  breaking  the 
beam  and  would  cause  twice  as 
much  deflection  as  if  it  was  loaded 
on  gradually.  A  railroad  train 
running  over  a  bridge  will,  for  the 
same  reason,  strain  the  bridge 
more  when  running  fast  than  it 
would  if  running  slow. 


FIG.  37 


To  Find  a   Suitable 


Size  of  Beam  for 
Deflection. 


a  Given  Limit  ol 


For  a  square  beam  supported  under  both  ends  and  loaded 
at  the  center,  use  the  formula : 


264  STRENGTH   OF   MATERIALS. 

4  '•'•• 

//"3   P  /• 

Side  of  the  beam  —  -i/±l-±_± 

o 

A  round  beam  supported  under  both  ends  and  loaded  at  the 
center  may  be  calculated  by  the  formula : 


Diameter  of  beam  =  J^-8  p  1-7  c 

O 

A  rectangular  beam  supported  under  both  ends  and  loaded 
at  the  center,  and  having  its  depth  1^  times  its  width,  may  be 
calculated  by  the  formula  : 


Depth  or  thickness  of  beam  = 


2S 

L  =  Length  of  span  in  feet. 

P  =  Center  load  in  pounds. 

S=  Given  deflection  in  inches. 
c  =  Constant  given  in  Table  No.  31. 

NOTE. — These  three  formulas  are  only  approximate,  as 
the  weight  of  the  beam  itself  is  not  considered ;  but  if 
necessary,  after  the  size  of  beam  is  obtained,  its  weight  may 
be  calculated  and  five-eighths  of  it  added  to  the  center  load, 
P\  and  using  the  same  formula  again,  another  beam  may 
be  calculated  for  this  new  center-load,  and  this  new  calculation 
will  give  a  beam  only  a  mere  trifle  too  small.  Constants  in 
Table  No.  31  are  for  beams  supported  under  both  ends  and 
loaded  at  the  center.  For  any  other  mode  of  loading  or  fasten- 
ing, constant  c  must  be  multiplied  according  to  rules  on 
page  261. 

To  Find  the  Constant  for  Deflection. 

If  experiments  are  made  upon  rectangular  beams,  use 
formula, 

SH*B 
L*  (P  +  ft  W) 
EXAMPLE. 

Calculate  the  constant  r,  or  deflection  in  inches  per  pound 
of  load,  for  a  beam  of  1  foot  span  and  1  inch  square,  supported 
under  both  ends  and  loaded  at  the  center,  when  experiments  are 
made  upon  a  pitch  pine  beam  40  feet  long,  12"  by  8'',  weighing 
1200  pounds  and  deflecting  1^  inches  for  a  center-load  of  500 
pounds. 

Solution : 

_  1.5  X  123  X  8 

403  X  (500  +  %  X  1200) 
c  =  0.000259  inch. 


STRENGTH  OF  MATERIALS.  265 

Modulus  of    Elasticity   Calculated  from   the  Transverse 
Deflection  in  a  Beam. 

When  experiments  are  made  upon  rectangular  beams  sup- 
ported under  both  ends  and  loaded  at  the  center,  the  modulus 
of  elasticity  may  be  calculated  by  the  formula, 


E  = 

±ST*B 

E  =  Modulus  of  elasticity. 
L  =  Length  of  span  in  inches  (not  in  feet). 
P  =  Load  in  pounds. 
W  =  Weight  of  beam  in  pounds. 
S  =  Deflection  of  beam  in  inches. 
T=  Thickness  of  beam  in  inches. 
B  =  Width  of  beam  in  inches. 

EXAMPLE. 

Calculate  the  modulus  of  elasticity  for  a  pitch  pine  rec- 
tangular beam  weighing  1200  pounds,  40  feet  span,  and  12"  by 
8",  deflecting  1)4  inches  for  a  center-load  of  500  pounds. 
(This  beam  and  conditions  are  the  same  as  mentioned  in 
the  previous  example  for  calculating  constants.) 

Solution  : 

E  _  4808X  (500  +#X  1200) 

4  X  ll/2  X  123  X  8 
E  _  138240000000 

82944 
E  —  1,666,666  pounds  per  square  inch. 

This  deflection  was  obtained  by  actual  experiments  on  a 
pitch  pine  beam  of  the  dimensions  given,  and  the  calculated 
modulus  of  elasticity  agrees  fairly  well  with  what  is  usually 
given  by  different  authorities  in  tables  of  modulus  of  elasticity. 
When  experimenting  it  is  necessary  to  take  the  average  of 
several  experiments  with  different  loads  and  to  try  the  beam  by 
turning  it  upside  down,  as  very  frequently  it  will  then  deflect  a 
different  amount  under  the  same  load.  Care  should  be  taken 
that  the  load  is  not  so  great  as  to  strain  the  beam  beyond 
its  elastic  limit.  As  long  as  the  deflection  increases  regularly  in 
proportion  to  the  load,  it  is  a  sign  that  the  elastic  limit  is  not 
reached.  It  is  very  difficult  to  ascertain  exactly  when  deflection 
will  commence  to  increase  faster  than  the  load,  because  material 
is  never  so  homogeneous  but  that  the  deflection  will  be  more  or 
less  irregular,  although  by  care  and  patience  fairly  good  re- 
sults may  be  obtained, 


266  STRENGTH   OF  MATERIALS. 

Allowable  Deflection. 

The  greatest  amount  of  deflection  which  may  be  allowed  in 
different  kinds  of  construction  can  only  be  determined  by  prac- 
tical experience  and  good  judgment  of  the  designer.  As  a  rule, 
in  iron  work  the  deflection  is  seldom  allowed  to  exceed  y^  of  the 
span,  which  is  equal  to  T^,  or  0.008  inch  per  foot  of  span.  Line 
shaftings  are  sometimes  allowed  to  deflect  j^  of  the  distance 
between  hangers  which  is  equal  to  0.01  inch  per  foot  of  span, 
but  head  shafts  carrying  large  pulleys  are  generally  not  allowed 
to  deflect  more  than  0.005  per  foot  of  span. 

In  woodwork,  considerable  more  deflection  is  allowed  than 
in  iron  structures.  Beams  in  houses  are  frequently  allowed  to 
deflect  -gfa,  or  even  T|¥  of  the  span;  this  is  equal  to  0.024  to 
0.025  inch*,  per  foot  of  span.  Woodwork  to  which  machinery  is 
to  be  fastened  must  never  be  allowed  to  deflect  so  much.  Such 
woodwork  must  always  be  so  stiff  that  it  supports  the  machinery, 
and  not  vice  versa;  for  instance,  in  beams  or  posts  by  which 
hangers  and  shafting  are  supported,  it  is  not  all-sufficient  that 
they  are  strong  enough,  but  they  must  also  always  be  stiff  enough. 

In  factories  it  is  very  important  that  floor  beams  as  well 
as  beams  supporting  heavy  shafting  have  sufficient  stiff- 
ness as  well  as  strength.  Floors  in  factories  are  frequently 
loaded  up  to  300  pounds  per  square  foot  of  surface.  For  floors 
in  public  buildings,  which  are  never  loaded  with  more  than  the 
weight  of  the  people  who  can  get  room,  the  load  will  hardly  ex- 
ceed 150  pounds  per  square  foot  of  surface.  Floors  in  tene- 
ment houses  are  seldom  loaded  more  than  60  pounds  per  square 
foot. 

Slate  roofs  weigh  about  8.5  pounds  per  square  foot  of 
surface.  Snow  may  be  reckoned,  when  newly  fallen,  to  weigh 
5  to  15  pounds  per  cubic  foot,  and  when  saturated  with  water 
it  may  weigh  40  to  50  pounds  per  cubic  foot.  Usual  practice  is 
to  allow  15  to  20  pounds  per  square  foot  for  snow  and  wind  on 
roofs. 

TORSIONAL  STRENGTH. 

The  fundamental  formula  for  torsional  strength  is, 


Pm  =  Twisting  moment,  and  is  the  product  of  the  length 
of  the  arm,  m,  in  inches  and  the  force,  /*,  in  pounds. 

S  —  Constant  computed  from  experiments,  and  is  some- 
times called  the  modulus  of  torsion  ;  its  value  usually  agrees 
closely  to  the  ultimate  shearing  strength  per  square  inch  of  the 
material. 

J  —  Polar  moment  of  inertia  (see  page  297). 

a  —  The  distance  in  inches  from  the  axis  about  which  the 
twisting  occurs  to  the  most  remote  part  of  the  cross  section. 

*  0.025  is  one-fortieth  inch  per  foot  of  span. 


STRENGTH    OF   MATERIALS.  267 


EXAMPLE  1. 


A  round  cast-iron  bar  3  inches  in  diameter,  is  exposed  to 
torsional  stress;  the  length  of  the  lever,  OT,  is  18  inches.  Find 
the  breaking  force,  P,  in  pounds  when  the  modulus  of  torsion 
for  cast-iron  is  taken  as  25,000  pounds. 

Polar  moment  of  inertia  for  a  circle  of  diameter,  </,  is, 


32 
The  distance  a  =  */*  d. 

25000  X  3*  X  3.1416  X 


_ 


18  X  1% 
25000  X  81  X  3.1416 


27  X  32 
pounds. 

The  advantage  of  the  above  formula  is  that  it  may  be 
used  for  any  form  of  section,  because  it  takes  in  the  polar 
moment  of  inertia  of  the  section;  but  it  is  seldom  that 
calculations  of  torsional  strength  are  required  for  other  than 
beams  of  round  or  square  section,  either  hollow  or  solid,  and 
the  strength  of  such  beams  may  be  more  conveniently  cal- 
culated in  an  equally  correct,  but  easier  way,  obviating  the  use 
of  both  the  polar  moment  of  inertia  and  the  modulus  of  torsion, 
by  reasoning  thus  : 

Consider  two  shafts,  a  and  £,  Fig  38.  Shaft  a  has  twice 
the  diameter  of  shaft  b,  and  consequently  four  times  the  area  ; 
therefore  it  has,  so  to  say,  four  times  as  many  fibers  to  resist 
the  stress,  and  for  this  reason  it  must  be  four  times  as  strong 
as  shaft  b  ;  but,  further,  the  outside  fibers  in  a  are  twice  as  far 
from  the  center,  therefore  the  fibers  in  shaft  a  must  also  have 
on  an  average  twice  the  advantage  over  the  fibers  in  shaft  b  to 
resist  the  twisting  effort  of  any  load  exerting  a  twisting  stress, 
and  for  this  reason  shaft  a  must  have  twice  the  strength  of  b  ; 
and  taking  these  two  reasons  together,  shaft  a  must,  conse- 
quently, be  eight  times  as  strong  as  shaft  £,  in  resisting  tor- 
sional stress. 

Thus,  the  strength  of  a  solid  shaft      /?%%ffi%fo>.       Flc'  38- 
increases  as  the  cube  of  the  diameter. 
Shaft  a  is  twice  as  large  in  diameter 
as  shaft  ft,  and  is,  therefore,  eight  times 
as  strong    as  £,   because  23  =  8.     If 
shaft  a  had  been  three  times  as  large 
as  b  it  would  have  been  27  times  as 
strong,  because  3s  =  27  ;  if  shafts  had 
been  four  times  as  large  as  £,  it  would  have  been  64  times  as 
strong,  because  43  =  64. 


268 


STRENGTH    OF   MATERIALS. 


Therefore,  if  the  constant  corresponding  to  a  load,  which, 
applied  to  an  arm  one  foot  long  will  twist  off  or  destroy  a  bar 
one  inch  in  diameter,  is  found,  the  breaking  load  for  any  round 
shaft  of  the  same  material  when  under  torsional  stress  may  be 
easily  calculated.  The  torsional  strength  (but  not  the  torsional 
deflection  in  degrees)  is  independent  of  the  length  of  the  shaft. 
The  strength  depends  only  upon  the  kind  and  the  amount  of 
material,  and  the  form  of  cross-section.  A  square  shaft  having 
its  sides  equal  to  the  diameter  of  a  round  shaft  will  have  ap- 
proximately 20%  more  strength  than  the  round  one,  but  it  will 
take  nearly  28%  more  material.  A  square  shaft  of  the  same 
area  as  a  round  shaft  has  approximately  15%  less  torsional 
strength  than  the  round  one. 

Thus: 

Formulas  for  torsional  strength  relating  to  solid  round 
shafts  will  be : 


Pm 


P  =  Breaking  load  in  pounds. 

D  =  Diameter  of  shaft  in  inches. 

;//  =  Length  in  feet  of  the  arm  on  which  load  P  is  acting. 
c  •=•  Constant,  and  it  is  the  load  in  pounds  which,  when 
applied  to  an  arm  one  foot  long,  will  twist  off  or  destroy  a  round 
bar  one  inch  in  diameter.  This  constant  is  obtained  from  ex- 
periments, and  is  given  in  Table  No.  32. 

RULE. — Multiply  the  cube  of  the  diameter  in  inches  by  the 
constants,  in  pounds,  divide  this  product  by  the  length  of  the 
lever  /«,  in  feet,  and  the  quotient  is  the  breaking  load  in  pounds. 

TABLE  No.  32.— Constant  c. 

The  ultimate  torsional  strength  in  pounds  of  a  round  beam 
one  inch  in  diameter,  when  load  is  acting  at  the  end  of  a  lever 
one  foot  long. 


MATERIAL. 

Very  Good. 

Medium  Good. 

Poor. 

Cast  Steel  

2  000 

1,000 

600 

Machinery  Steel*  .  .  . 
Wrought  Iron  .... 
Cast-iron 

1,200 

800 
525 

1,100 
580 
450 

700 
500 
350 

*  Machinery  steel  or  wrought  iron  may  not  actually  break  at  this  load,  but  it 
will  deflect  and  yield  so  it  will  become  useless. 


STRENGTH  OF  MATERIALS.  269 

EXAMPLE  1. 

A  wrought  iron  shaft  is  eight  inches  in  diameter,  and  the 
force  acts  upon  a  lever  two  feet  long.  How  much  force  must 
be  applied  in  order  to  twist  off  or  to  destroy  the  shaft? 

Solution : 

P  _  83  X  580     _  512  X  580 
2  ~2~ 

EXAMPLE  2. 

A  force  of  870  pounds  is  acting  with  a  leverage  of  four  feet 
in  twisting  a  wrought  iron  shaft.  What  must  be  the  diameter 
of  the  shaft  in  order  to  resist  the  twisting  stress,  with  10  as  a 
factor  of  safety? 

Solution : 

3 

r^  _  .  /  P  *n  X  10 


X  4  X  10 

580 

D  —  V60"  =  3.914,  or,  practically,  a  4-inch  shaft. 

NOTE. — Ten  is  used  as  a  multiplier  of  the  twisting  moment, 
P  m,  because  10  is  the  factor  of  safety.  Constant  580  is  taken 
from  Table  No.  32. 

EXAMPLE  3. 

A  round  bar  of  cast-iron  four  inches  in  diameter  is  to  be 
twisted  off  by  a  force  of  3200  pounds.  How  long  a  leverage  is 
necessary  ?  (c  for  cast-iron,  in  Table  No.  32,  is  450). 

Solution : 


EXAMPLE  4. 

Experiments  are  made  upon  a  cast-iron  round  bar  2  inches 
in  diameter  with  a  leverage  of  5X  feet ;  the  bar  is  twisted  off  at 
a  force  of  832  pounds.  Calculate  constant  c,  or  the  force  in 
pounds  if  acting  with  a  leverage  of  one  foot,  which  will  break  a 
round  bar  of  the  same  material  one  inch  in  diameter. 

Solution : 

'^TT 

^882X5X^4368   =  540  pounc,5. 


270  STRENGTH    OF    MATERIALS. 

Hollow  Round  Shafts. 

In  proportion  to  the  amount  of  material  used,  around  hollow 
shaft  has  more  torsional  strength  than  a  solid  shaft  of  the  same 
diameter.  This  is  because  the  fibers  in  any  shaft  exposed  to 
twisting  stress  only  offer  resistance  to  the  load  in  proportion  to 
their  stretch.  Therefore,  the  fibers  near  the  center  are  always 
in  position  to  offer  less  resistance  than  the  fibers  more  remote 
from  the  center. 

The  formula  for  torsional  strength  in  round  hollow  shafts 
will  be  : 

—  d* 


P  —  Ultimate  breaking  load  in  pounds  applied  at  a  leverage 

of  m  feet. 
D  —  Outside  diameter  of  shaft  in  inches. 

d=  Inside  diameter  of  shaft  in  inches. 
m  —  Length  of  lever  in  feet. 

c  =  Constant  (same  as  for  a  solid  shaft). 


Square  Beams  Exposed  to  Torsional  Stress. 

The  theoretical  formula  for  twisting  strength  (on  page  26(5) 
will  apply  to  square  as  well  as  round  beams.  The  proportional 
strength  between  a  round  and  a  square  beam  may,  therefore,  be 
compared  by  using  that  formula.  Let  S  represent  the  side  of  a 
square  beam  and  the  polar  moment  of  inertia  is  J  S*. 

The  distance  from  the  center  of  the  beam  to  the  most 
remote  fiber  in  a  square  beam  is  S  -v  %,  and,  dividing  the  polar 
moment  of  inertia  by  this  distance,  we  have, 


•sV  y2 

Let  D  represent  the  diameter  of  a  round  beam.  The 
polar  moment  of  inertia  is  ±_!  =  0.098  LP 

The  distance  from  the  center  to  the  most  remote  fiber  in  the 
round  beam  is  l/2  D.     Dividing  the  polar  moment  of  inertia  by 

this  distance,  we  have  °-Q98  ^  —  0.195  D* 

Suppose,  now,  that  S  and  D  are  equal,  for  instance. 
one  inch  ;  the  proportion  in  torsional  strength  between  the  two 
beams  must  be  0.23  divided  by  0.19G,  which  equals  1.18. 
Thus,  for  square  beams,  use  the  formulas  given  for  round 
beams,  but  multiply  constant  <:,  in  Table  No.  32,  by  1.2,  and 


STRENGTH   OF  MATERIALS.  271 

take  the  side  instead  of  the  diameter.     The  formula  for  tor- 
sional  strength  in  a  square  beam  will  be  : 
P  _  (  Side  )3  X  1.2  X  c 
Length  of  leverage. 

c  =  Constant  (same  as  for  a  round  beam). 

P=  Load  in  pounds. 

Side  is  measured  in  inches. 

Length  of  leverage  is  measured  in  feet. 

Torsional  Deflection. 

The  torsional  deflection  in  degrees  will  increase  directly 
with  the  length  of  the  shaft  and  the  twisting  load,  and  inversely 
as  the  fourth  power  of  the  diameter  of  the  shaft ;  therefore,  the 
formula  for  torsional  deflection  is  : 
?_c  X  m  X  L  X  P 

—D^~ 

S  =  Deflection  in  degrees  for  the  length  of  the  shaft. 
m  =  Length  of  lever  in  feet. 
L  =  Length  of  shaft  in  feet. 
P  =  Load  in  pounds. 
D  =  Diameter  of  shaft  in  inches. 

c  =  Constant  obtained  from  experiments  for  different 
kinds  of  material,  and  is  the  deflection  in  degrees  for  a  shaft 
one  inch  in  diameter  and  one  foot  long,  when  loaded  with  one 
pound  on  the  end  of  a  lever  one  foot  long. 

The  author  of  this  book  has  made  experiments  on  tor- 
sional deflection  in  wrought  iron  shafts  two  inches  in  diameter. 
The  average  deflection  was  ll/2  degrees  in  10  feet  of  length, 
when  a  load  of  50  pounds  was  applied  on  a  lever  5#  feet  long. 
Constant  <:,  as  calculated  from  these  experiments,  will  be  0.00914. 
Using  this  constant,  the  formula  for  torsional  deflection  for 
wrought  iron  will  be  : 

9  —  L  x  m  x  p  x  °-00914 
D* 

Machinery  steel  and  wrought  iron  will  deflect  about  the 
same.  Cast-iron  will  deflect  twice  as  much  as  wrought  iron.  A 
square  bar  will  deflect  0.589  times  as  much  as  a  round  bar  when 
side  and  diameter  are  alike. 

Formula  for  Torsional  Deflection  in  Hollow  Round  Shafts. 

~_L  X  m  X  PXc 
Lfi-d* 

D  =  Outside  diameter  in  inches. 
d=  Inside  diameter  in  inches. 

All  the  other  letters  have  the  same  meaning  as  explained 
under  formulas  for  solid  shafts. 


272 


STRENGTH  OF    MATERIALS. 

TABLE  No.  33.— Constant  c. 


The  torsional  deflection  in  degrees  per  foot  of  length  for  a 
shaft  of  one  inch  side  or  diameter  when  loaded  with  one  pound 
at  the  end  of  a  lever  one  foot  long : 


MATERIAL. 

Round 
Section. 

Square 
Section. 

Machinery  Steel              .   . 

0  00914° 

0  00538° 

\Vrought  Iron 

0  00914° 

0  00538° 

Cast-iron                                .       • 

0  018° 

0  0106° 

Oak  

0.795° 

0.468° 

Ash   .    

0.784° 

0.460° 

Pine  and  Spruce 

1  35° 

0  79° 

EXAMPLE. 

A  round  bar  of  wrought  iron  16  feet  long  and  3  inches  in 
diameter  is  fastened  at  one  end  and  the  other  is  exposed  to  a 
twisting  load  of  1000  pounds,  acting  with  5  feet  leverage.  How 
many  degrees  will  this  load  deflect  the  bar  ? 

Solution : 

16  X  5  X  1000  X  0.00914 


S= 


34 


S= 


731.2 

81 


S=  9  degrees. 

NOTE. — From  Table  No.  33,  it  is  seen  that  only  steel  and 
wrought  iron  are  suitable  for  shafts  exposed  to  torsional  stress. 
Wrought  iron  is  about  twice  as  good  as  cast-iron,  over  80  times 
better  than  oak,  and  about  150  times  as  good  as  pine. 


SHEARING    STRENGTH. 

Sometimes  force  may  act  in  such  a  manner  that  the 
material  is  sheared  off.  For  instance,  the  rivets  in  a  steam 
boiler  are  exposed  to  shearing  stress  (see  Fig.  39)  when  the 
boiler  is  under  steam  pressure. 

FIG.  39. 

When  holes  are  punched  or  bars  of      j— 

iron  are  cut  off  under  punching  presses,      ' 

the  action  of  the  punch  in  cutting  off  the      -« •»  »»• »- 

material  is  shearing,  and  the  resistance  which  the  material  offers 
is  its  ultimate  shearing  strength.  The  average  ultimate  shear- 
ing strength  of  wrought  iron  is  40,000  pounds  per  square  inch, 


STRENGTH   OF  MATERIALS.  273 

In  cast-iron  the  ultimate  shearing  strength  is  usually  between 
20,000  and  30,000  pounds  per  square  inch.  In  steel  the  ultimate 
shearing  strength  will  vary  from  40,000  to  80,000  pounds  per 
square  inch. 

The  resistance  offered  to  shearing  is  in  proportion  to 
the  sheared  area.  Thus,  it  will  take  twice  as  much  force 
to  punch  a  hole  two  inches  in  diameter  through  a  three-eighths 
inch  plate  as  it  would  to  punch  a  hole  only  one  inch  in  diameter 
through  the  same  plate,  and  it  will  take  four  times  as  much 
force  to  shear  off  a  one-inch  bolt  as  it  would  to  shear  off  a  one- 
half  inch  bolt,  because  the  area  of  a  one-inch  bolt  is  four  times 
as  large  as  the  area  of  a  one-half  inch  bolt. 

EXAMPLE  1. 

How  much  force  is  required  to  shear  off  a  wrought  iron 
rivet  of  one-inch  diameter  if  the  shearing  strength  of  the 
wrought  iron  is  40,000  pounds. 

Solution : 

One-inch  diameter  —  0.7854  square  inches;  therefore  the 
force  required  will  be  0.7854  X  40,000  —  31,416  pounds. 

EXAMPLE  2. 

A  wrought  iron  plate  is  one-quarter  of  an  inch  thick  and 
the  ultimate  shearing  strength  of  the  iron  is  40,000  pounds  per 
square  inch.  How  much  pressure  is  required  to  punch  a  hole 
three-quarters  of  an  inch  in  diameter  ? 

Solution : 

The  circumference  of  a  ^-inch  circle  is  2.356  inches.  The 
plate  is  X-inch  thick;  therefore  the  area  of  shearing  surface, 
2.3562  X  X  =  0.58905 ;  thus,  the  force  required  will  be  40,000 
X  0.58905  =  23,562  pounds. 


TABLE  No.  34.— Shearing  Strength  Per  Square  Inch. 


MATERIAL. 

Pounds  Per  Square  Inch. 

Steel         .                               

45  000  to  75  000 

Wrought  Iron  Rivets              

35  000  to  55  000 

Cast-iron   •  

20000  to  30000 

Oak,  crosswise    

4500  to    5500 

400  to        700 

4,000  to     5  000 

Pitch  Pine  lengthwise                        • 

400  to        600 

Spruce   crosswise                     • 

3000  to     4000 

Soruce.  lengthwise 

300  to        500 

274 


STRENGTH    OF    MATERIALS. 

FACTOR  OF  SAFETY. 


The  factor  of  safety  can  only  be  fixed  upon  by  the  experi- 
ence and  good  judgment  of  the  designer.  It  may  vary  from  4  to 
40.  In  a  temporary  structure,  when  the  greatest  possible  load  to 
which  it  will  be  exposed  is  known,  a  factor  of  safety  of  four 
may  be  safe  enough,  but  frequently  a  greater  factor  is  necessary. 
Different  factors  of  safety  are  also  necessary  for  different 
materials;  a  different  factor  of  safety  may  also  be  necessary 
in  different  parts  of  the  same  machine.  The  following  Table, 
No.  35,  is  only  offered  as  a  guide  in  selecting  factor  of  safety : 

TABLE  No.  35.— Factor  of  Safety. 


Dead  Load, 

Variable  Load, 

Machinery 

MATERIAL. 

such  as  build- 
ings contain- 
ing little  or  no 

such  as  bridges 
and  slow- 
running 

Machinery 
in 
General. 

Exposed  to 
hard  usage,  as 
Rolling  Mills, 

machinery. 

machinery. 

etc. 

Steel, 

5 

7 

10 

15 

Wrought  Iron, 

4 

6 

10 

15 

Cast-iron, 

6 

10 

15 

25 

Brickwork, 

15 

25 

30 

40 

Wood, 

8 

10 

15 

20 

If  a  structure  is  exposed  to  stress  alternately  in  one  direc- 
tion and  then  in  another,  it  is  necessary  to  use  a  higher  factor  of 
safety  than  if  it  is  only  exposed  to  a  steady  stress  one  way.  A 
comparatively  small  load,  when  applied  a  sufficient  number  of 
times,  may  break  a  structure  or  a  machine,  although  it  does  not 
break  it  the  first  time.  For  instance,  commence  to  hammer  on 
a  bar  of  cast-iron  and  it  will  break  after  several  blows, 
although  the  last  blow  need  not  be  any  more  powerful 
than  the  first  one.  It  is  the  same  way  with  anything  else;  it 
may  break  in  time,  although  it  is  strong  enough  to  resist  the 
stress  at  the  beginning;  therefore,  within  practical  limits,  the 
larger  the  factor  of  safety  the  longer  time  the  structure  may 
last. 


NOTES  ON  STRENGTH  OF  flATERIAL. 

In  steel,  the  crushing  strength  usually  exceeds  the  tensile 
strength,  but  wrought  iron  has  usually  a  little  more  tensile  than 
crushing  strength,  and  its  shearing  strength  is  about  80  per  cent, 
of  its  tensile  strength.  Both  steel  and  wrought  iron  are  suitable 
to  resist  any  kind  of  stress,  and  compared  to  other  materials 
they  are  especially  adapted  for  anything  exposed  to  twisting  and 
shearing  stress. 


STRENGTH    OF   MATERIALS.  275 

Cast-iron  is  variable ;  it  has  usually  five  to  six  and  a-half 
times  as  much  crushing  as  tensile  strength,  and  when  loaded 
transversely  it  will  deflect  under  the  same  load  nearly  twice  as 
much  as  wrought  iron.  It  is  especially  useful -for  short  pillars 
or  anything  exposed  to  crushing  stress,  where  there  is  little 
danger  of  breakage  by  flexure;  it  is  very  much  less  reliable 
when  exposed  to  tensile  or  torsional  stress. 

Wood  is  not  adapted  to  resist  torsion,  but  is  useful  to  resist 
tensile,  crushing  and  transverse  stress,  also  to  resist  flexure. 
It  has  nearly  twice  as  much  tensile  as  crushing  strength ,;  there- 
fore, it  would  seem  specially  well  adapted,  in  all  kinds  of  con- 
struction, to  be  the  member  exposed  to  tensile  stress,  but  where 
wood  and  iron  enter  into  construction  together,  iron  is 
always  used  as  the  member  to  take  the  tensile  stress  and  wood 
as  the  compressive  member,  because  wood  has  suck  low  shear- 
ing strength  lengthwise  with  its  fibers  that,  with  any  kind  of 
fastening  at  the  ends,  it  will  tear  and  split  at  the  holes  under 
comparatively  little  stress;  but  this  difficulty  is  easily  overcome 
when  wood  is  used  as  the  compressive  member.  Wood  has 
comparatively  low  tensile  and  crushing  strength  crosswise  on  the 
fiber.  This  is  well  to  remember  with  beams  loaded  transversely 
and  laid  on  posts.  The  beams  may  be  sufficiently  strong,  but 
under  heavy  load,  if  suitable  precautions  are  not  taken  (see 
page  250)  the  top  of  the  post  may  press  into  the  beam,  especially 
if  the  lumber  is  green. 

Stone  has  high  crushing  strength  but  low  tensile  strength, 
and,  in  consequence,  very  low  transverse  strength.  It  is  very 
well  adapted  for  foundations  when  supported  and  laid  in  such  a 
way  that  its  crushing  strength  comes  into  play,  but  when  laid  as 
a  beam  to  resist  transverse  stress  it  is  very  unreliable,  as  it  will 
break  for  a  comparatively  small  load  and  it  may  break  from  a 
blow  or  jar. 

Brickwork  is  only  suitable  for  crushing  stress,  and  there  is 
great  difference  in  the  strength  of  different  kinds  of  brick. 

In  calculating  strength  and  stiffness  in  any  kind  of  design- 
ing, it  should  be  remembered  that  it  is  only  possible  to  deter- 
mine the  strength  of  any  material  by  actual  test,  and  that  the 
tabular  and  constant  numbers  here  given  are  only  an  average 
approximate. 


Mechanics, 


The  science  which  treats  of  the  action  offerees  upon  bodies 
and  the  effect  they  produce  is  called  Mechanics. 

Newton's  Laws  of  Motion. 

The  three  fundamental  principles  of  the  relation  between 
force  and  motion  were  first  stated  by  Sir  Isaac  Newton,  and  are 
therefore  called  Newton's  laws  of  motion. 

NEWTON'S  FIRST  LAW. 

All  bodies  continue  in  a  state  of  rest  or  of  uniform  motion 
in  a  straight  line,  unless  acted  upon  by  some  external  force  that 
compels  change. 

NEWTON'S  SECOND  LAW. 

Every  motion  or  change  of  motion  is  proportional  to  the 
acting  force,  and  the  motion  always  takes  place  in  the  direction 
of  a  straight  line  in  which  the  force  acts. 

NEWTON'S  THIRD  LAW. 

To  every  action  there  is  always  an  equal  and  contrary  re- 
action. 

Gravity. 

The  natural  attraction  of  the  earth  on  everything  on  its 
surface  which  will  cause  any  body  left  free  to  move  to  fall  in  the 
direction  of  the  center  of  the  earth  is  called  the  force  of 
gravity. 

Acceleration  Due  to  Gravity. 

If  a  body  is  left  free  to  fall  from  a  height,  its  velocity  will 
not  be  constant  throughout  the  whole  fall,  but  it  will  increase  at 
a  uniform  rate.  It  is  this  uniform  increment  in  velocity  which 
is  called  acceleration  of  gravity.  It  is  usually  reckoned  in  feet 
per  second.  A  body  falling  free  will  at  the  end  of  one  second 
have  acquired  a  velocity  of  32M$  feet,  or,  practically,  32.2  feet 
per  second;  but  it  has  fallen  through  a  space  of  16.1  feet, 
because  it  started  from  rest  and  the  velocity  was  increasing  at  a 
uniform  rate  until,  at  the  end  of  the  second,  it  was  32.2  feet  per 
second ;  therefore,  the  average  velocity  during  the  first. second 
can  only  be  16.1  feet.  At  the  end  of  two  seconds  the  velocity 
has  increased  to  64.4  feet  per  second  and  the  space  fallen 

(276) 


MECHANICS.  277 

through  is  64.4  feet,  because  the  average  velocity  per  second 
must  be  half  of  the  final  velocity  ;  therefore,  the  average 
velocity  is  32.2  feet  per  second,  and,  as  the  time  is  two  seconds 
the  space  will  be  04.4  feet.  At  the  end  of  three  seconds  the 
final  velocity  has  increased  to  3  X  32.2  —  96.6  feet  per  second 
and  the  space  fallen  through  is  -9  |  ii  x  3  =  144.9  feet,  etc.  This 
is  supposing  the  body  was  falling  freely  in  vacuum,  but  while 
the  air  will  offer  a  resistance  and  somewhat  reduce  the  actual, 
motion,  the  principle  is  the  same.  Acceleration  due  to  gravity 
varies  but  little  at  different  latitudes  of  the  earth.  At  the 
equator  it  is  calculated  to  be  :J2.088  and  at  the  pole  32.253  feet. 
Acceleration  due  to  gravity  decreases  at  higher  altitudes,*  but 
all  these  variations  on  the  earth's  surface  are  so  small  that  they 
hardly  need  to  be  considered  in  any  calculation  concerning 
practical  problems  in  mechanics. 

Velocity. 

The  velocity  of  falling  bodies  increases  at  a  uniform  rate  of 
32.2  feet  per  second  ;  therefore,  when  commencing  from  rest,  the 
final  velocity  in  feet  per  second  must  be, 


RULE. 

Multiply  the  time  in  seconds  by  32.2  and  the  product  is  the 
final  velocity  in  feet  per  second  ;  or,  multiply  the  height  of  the 
fall  in  feet  by  64.4  and  the  square  root  ot  the  product  is  the 
velocity  in  feet  per  second. 

EXAMPLE. 

What  final  velocity  will  a  body  acquire  in  a  free  fall  during 
seven  seconds  ? 

Solution  : 

if  —  7  X  32.2  =  225.4  feet  per  second. 

Height  of  Fall. 

The  average  velocity  per  second  is  always  half  of  the  final 
velocity  per  second.  Therefore  the  space  fallen  through  in  a 
given  time  is  found  by  multiplying  half  of  the  final  velocity  by 
the  number  of  seconds  which  produced  that  velocity.  Thus,  the 
formulas  : 

h  =  f-^-  =  tQ.$v=  v  0.5  /=  -?!  =  ££= 
2  2*2 

*  Above  the  surface  of  the  earth  the  weight  of  a  body  is  inversely  propor. 
tional  to  the  square  of  its  distance  from  the  center  ot  the  earth. 

Below  the  surface  of  the  earth  the  weight  of  a  body  is  directly  proportional  to 
its  distance  from  the  center  ot  the  earth. 


278  MECHANICS. 

EXAMPLE. 

A  fly-wheel  has  a  rim  speed  of  48  feet  per  second.     From 
hat  height  must  a  body  drop  to  acquire  the  same  velocity? 


w 

Solution  : 


=  =  85.78  feet. 


2  g         64.4  644 

Time. 

RULE. 

Divide  the  space  by  16.1,  and  the  square  root  of  the 
quotient  is  the  time  ;  or,  divide  given  velocity  by  32.2,  and  the 
quotient  is  the  time. 

r 


S          *  0.5  g 
EXAMPLE. 

How  long  a  time  does  it  take  before  a  body  in  a  free  fall 
acquires  a  velocity  of  100  feet  per  second  ? 
Solution : 

,-JL=  ^°  =  3.1  seconds. 
g          32.2 

Distance  a  Body  Drops  During  the  Last  Second. 

The  space  through  which  a  body  will  drop  in  the  last 
second  is  equal  to  the  final  velocity  minus  half  of  acceleration 
due  to  gravity.  Therefore,  this  space  is  found  by  the  formula: 

x  =  v  —  y2g  =  g(t  —  y2) 

x  =  Space  in  feet  which  the  body  drops  the  last  second  of 

the  fall. 

/  =  Time  in  seconds. 
v  =  Final  velocity. 

g  =  Acceleration  of  gravity  =  32.2  feet. 
h  —  Height  of  fall  in  feet. 
EXAMPLE. 

A  body  has  in  a  free  fall  obtained  a  final  velocity  of  40  feet 
per  second.     What  space  did  it  drop  the  last  second  ? 
Solution: 
x  —  v  —  y2g—  40  —  ^L2  =  40  —  16.1  =  23.9  feet. 

EXAMPLE. 

A  body  was  falling  four  seconds.     How  many  feet  did  it 
drop  the  last  second  ? 
Solution  : 
x  —  g(t  -  */2)  =  32.2  X  (4  —  #)  =  32.2  X  3.5  =  1 12.7  feet. 


MECHANICS. 


279 


TABLE  No.  35.— Time,  Velocity  and  Height. 

£•=32.161  Feet. 


Time  in  Seconds. 

Velocity  in  Feet  at 
the  End  of  the  Time. 

Height  of  Fall  in 
Feet. 

Distance  in  Feet  that 
the  Body  Drops  in 
the  Last  Second. 

1 

32.161 

16.08 

16.08 

2 

64.322 

64.32 

48.24 

3 

96.483 

144.72 

80.40 

4 

128.644 

257.28 

112.56 

5 

160.805 

402.00 

144.72 

Upward  Motion. 

A  body  thrown  perpendicularly  upward  with  a  certain 
velocity  will  continue  the  upward  movement  until  it  reaches  the 
same  height  from  which  it  would  have  to  fall  in  order  to  get  a 
final  velocity  equal  to  the  starting  velocity.  Therefore,  a  body 
projected  upward  with  a  given  velocity  will  return  again  witn 
the  same  velocity.  This  is  theoretical  in  a  vacuum,  but  actually 
the  body  neither  continues  to  the  theoretical  height  nor  returns 
with  a  final  velocity  equal  to  the  starting  velocity,  because  the 
air  will  always  offer  considerable  resistance.  The  greater  the 
weight  of  a  body,  in  proportion  to  its  volume,  the  nearer  the 
velocity,  when  it  returns,  will  be  equal  to  its  starting  velocity. 

EXAMPLE. 

A  body  is  projected  upward  with  a  velocity  of  45  feet  per 
second.  How  high  will  it  go  before  it  stops  and  commences  to 
drop  again,  the  resistance  of  the  air  not  being  considered  ? 

The  solution  of  this  problem  is  simply  to  find  the  theoretical 
height  from  which  a  body  must  drop  to  attain  a  final  velocity  of 
45  feet,  which  is  solved  by  the  formula, 


*-£- 


2025 
64.4 


=  31.286  feet. 


Body  Projected  at  an  Angle. 

If  a  body  is  projected  in  the  direction  of  the  line  d  e  (see 
Fig.  1),  with  an  initial  velocity  per  second  equal  to  the  distance 
from  d  to  1.  no  force  acting  after  the  body  is  started,  it  will  con- 
tinue to  move  at  constant  velocity  in  a  straight  line  indefinitely ; 
at  the  end  of  the  first  second  it  would  be  at  1,  at  the  end  of 
two  seconds  it  would  be  at  2,  at  the  end  of  the  third  second  at  3, 
at  the  end  of  the  fourth  second  at  4,  etc.;  but,  on  account  of  the 
force  of  gravity,  the  motion  will  be  entirely  different.  The 
force  of  gravity  acts  on  this  body  exactly  as  if  it  was  falling  in 


280 


MECHANICS. 


a  vertical  line.  At  the  end  of  the  first  second  the  force  of 
gravity  has  caused  this  moving  body  to  drop  16.1  feet  out  of  its 
path;  therefore,  instead  of  being  at  1  at  the  end  of  the  first 
second,  it  is  at  a  point  16.1  feet  vertically  under  1 ;  instead  of 
being  at  2  at  the  end  of  two  seconds,  it  is  at  a  point  2  X  2  X 
16.1  =  64.4  feet  vertically  below  2;  instead  of  being  at  3  at  the 
end  of  the  third  second,  it  is  at  a  point  3  X  3  X  16.1  =  144.9 
feet  vertically  below  3 ;  and  instead  of  being  at  4  at  the  end  of 
the  fourth  second,  it  is  at  a  point  4  X  4  X  16.1  =  257.6  feet 
vertically  below  4,  etc. 


FIG.  1. 


When  a  body  is  projected  in  a  vertical  upward  direction 
with  an  initial  velocity  of  v  feet  per  second,  it  proceeds  to  a 

•z/2 
height  ~2~zr;  therefore,  when  projected  at  an  angle,  a  (see  Fig. 

1),  with  a  velocity  of  i>  feet  per  second,  it  will  proceed  to  the 

?/2  sin.2  a 
height 2^— 

When  a  body  is  projected  in  a  vertical  upward  dirction  with 
a  velocity  of  v  feet  per  second,  the  time  for  ascent  is  — -  and  the 

time   for  descent  is  equal  to  the  time  for  ascent;    therefore, 

v 

the  total   time   will  be  2 — ;  but  when  the  body  is  projected 

upward  at  an  angle  of  a  degrees,  the  total  time  for  ascent  and 

2  v  sin.  a 
descent  will  be 

o 

The  horizontal  distance,  or  the  range  from  d  to  n,  will  be 
equal  to  the  velocity  in  feet  per  second  multiplied  by  the  total 


MECHANICS.  28l 

number  of  seconds  consumed  in  the  ascent  and  descent,  and 
this  multiplied  by  cos.  of  the  angle  a;  therefore, 

/2  T/ sin.  a\  2  7/2  sin.  a  cos. a 

Horizontal  range  =  v  ^ )  cos.  a  =  —         ——_ — 

but  2  X  sin.  a  X  cos.  a  is  always  equal  to  sin.  of  an  angle  of 
twice  as  many  degrees  as  the  angle  a.    Therefore,  the  formula 

•z/2  sin.  2  a 
Deduces  to  horizontal  range  —      — — — 

Thus,  the  following  formulas  will  apply  to  bodies  projected 
at  an  angle.  (  See  Fig.  1). 

The  greatest  possible  height  will  be, 

*  =  **£• 

The  greatest  possible  range! will  be, 

g 

The  time  in  seconds  will  be, 

t_  2  v  sin,  a  _ 

g 

v  •=.  Velocity  in  feet  per  second. 
g  =  Acceleration  of  gravity  =  32.2. 

TO  FIND  THE  HEIGHT  TO  WHICH  A  BODY  CAN  ASCEND. 

RULE. 

Multiply  the  velocity  in  feet  per  second  by  the  sine  of  the 
angle  (to  the  horizontal  line),  square  this  product  and  divide 
by  64.4,  and  the  quotient  is  the  height  in  feet. 

TO  FIND  THE  LONGEST  POSSIBLE   RANGE. 

RULE. 

Multiply  the  square  of  the  velocity  in  feet  per  second  by 
sine  of  an  angle  of  twice  as  many  degrees  as  the  angle  of  the 
throw  (to  the  horizontal  line),  and  divide  by  32.2.  The 
quotient  is  the  longest  distance  the  body  can  be  thrown. 

TO  FIND  THE  TIME  OF  FLIGHT. 

RULE. 

Multiply  the  velocity  in  feet  per  second  by  sine  of  the 
angle  (to  the  horizontal  line),  and  divide  by  16.1.  The 
quotient  is  the  time  in  seconds. 

EXAMPLE. 


of  55°  to  the  horizontal 
line,  with  an  initial  velocity  of  120  feet  per  second.    How  high 


A  body  is  projected  at  an  angle 
j,  with  an  initial  velocity  of  120  fee 


282  MECHANICS. 

will  it  go  ?     How  far  will  it  go  in  a  horizontal  direction  ?     How 
many  seconds  will  it  take  to  finish  the  flight? 
Solution  for  height  : 

T        2/2  sin.2  a 

11  --     - 


^  _  1202  X  sin.2  55° 

h  = 


-644— 
1202  X  0.819152 


64.4 
h  _  14400  X  0.673 

d4.4 

h  —  150.5  feet. 
Solving  for  horizontal  range : 
^  __  -z/2  sin.  2  a 
g 

Twice  the  angle  of  55°  is  110°  and  sine  of  110°  will  be  sine 
of  70°,  because  180°  —  110°  =  70° ;  therefore,  sine  of  110°  equals 
sine  of  70°  in  the  second  quadrant,  and  the  solution  will  be  : 

1202  X  sin.  70° 


b  = 


32.2 


r         14400  X  0.93969 

322         "  =  128'4  feet 


Solving  for  time  of  flight: 


0.5  g 

.  _  120  X  sin.  55° 
16.1 

.  _  120  X  0.81915 

*-         — jTTj —    -  =  6.1  seconds. 

EXAMPLE. 

A  nozzle  on  a  hose  is  placed  at  an  angle  of  28°  to  the 
horizontal  line  and  the  spouting  water  when  leaving  the  nozzle 
has  a  velocity  of  36  feet  per  second.  How  far  will  it  theoretic- 
ally reach  in  a  horizontal  direction? 

Solution : 

Range  =  b  =  *  sin'  2  * 

g 
362  X  sin.  56° 


b  = 

g 
,   _  1296  X  0.82904 


feet' 


MECHANICS.  283 

EXAMPLE  3. 

A  nozzle  on  a  hose  is  placed  at  an  angle  of  38°  to  the 
horizontal  line  and  is  spouting  water  a  distance  of  40  feet  in  a 
horizontal  direction.  What  is,  theoretically,  the  velocity  of  the 
water  when  leaving  the  nozzle  ? 

Solution: 


sin.  76 


—  _  (40  X  o2.2  o/j  A  r~~i.  ,^«  sprond 


0.9703 

NOTE. — In  Example  2  we  multiply  by  sine  of  56  degrees, 
because  water  is  leaving  the  nozzle  at  an  angle  of  28  degrees, 
and  twice  28  equals  56.  In  Example  3  we  multiply  by  sine  of  76 
degrees,  because  twice  38  equals  76.  See  previous  explanations. 

The  greatest  possible  height  will  be  reached  if  the  body  is 
thrown  perpendicularly  upward.  The  greatest  possible  range  is 
obtained  if  the  body  is  thrown  at  an  angle  of  45°  and  will  then  be  : 

"  ~g~ 

At  an  angle  of  45°  the  horizontal  range  will  be  twice 
the  greatest  possible  height  which  could  have  been  reached  if 
the  body  had  been  thrown  perpendicularly  upward.  At  this 
angle  the  horizontal  range  is  four  times  the  height.  For  an 
equal  number  of  degrees  over  or  under  45  degrees  the  horizontal 
range  will  be  equal ;  for  instance,  if  a  body  is  thrown  out  at  an 
angle  of  30  or  60  degrees,  the  horizontal  distance  is  the  same, 
but  the  height  of  ascension  will  be  much  more  at  60  degrees 
than  at  30  degrees.  It  is  frequently  useful  to  notice  this  in 
practical  work.  For  instance,  water  under  pressure  is  thrown 
the  farthest  distance  in  a  horizontal  direction  from  a  hose  when 
the  nozzle  is  held  at  an  angle  of  45  degrees  to  the  horizontal 
line.  It  is  possible  by  the  same  pressure  to  throw  water  twice 
as  far  in  a  horizontal  distance  as  in  vertical  height. 

Motion  Down  an  Inclined  Plane. 

A  ball  rolling  along  an  incline,  as  n  f-*^  F|G-  2. 
a  c  (Fig.  2),  will  have  the  same 
velocity  when  it  gets  to  c  as  it  would 
have  had  if  dropping  freely  from  a 
to  £,  supposing  all  friction  to  be  left 
out  of  consideration. 

The  average  velocity  will  also  be 
half  of  the  final  velocity,  and  the 

time  used  in  the  fall  will  be  the  distance  a  c  (the  length  of  the 
incline),  divided  by  the  average  velocity  per  second. 


284  MECHANICS. 

Body  Projected  in  a  Horizontal  Direction  Prom  an 
Elevated  Place. 

When  a  body  is  projected  in  a  horizontal  direction  from  a 
place  which  is  higher  than  the  one  where  it  strikes  the  ground, 
the  range  in  feet  in  a  horizontal  direction  will  be  equal  to  the 
product  of  velocity  in  feet  per  second  and  the  time  in  seconds 
which  it  will  take  for  a  body  in  a  free  fall  to  drop  a  distance 
equal  to  the  difference  in  vertical  height  between  the  two 
places.  Thus  : 


VO~T 
~JL 
s 

•v  =  Initial  velocity  in  feet  per  second. 

h  =  Vertical  height  in  feet. 

g=  Acceleration  of  gravity  =  32.2  feet. 


EXAMPLE. 

Water  spouts  from  a  nozzle  in  a  horizontal  direction  at  a 
velocity  of  30  feet  per  second  and  the  nozzle  is  placed  12  feet 
above  the  ground.  What  is  the  horizontal  range  of  the  water  ? 

Solution  : 

Horizontal  range  =  v  -\j—  =  30  -^12  X2  =  22.45  feet. 


To  Calculate  the  Speed  of  a   Bursted  Fly-Wheel  from 
the  Distance  the  Fragments  are  Thrown. 

The  angle  of  45  degrees  is  the  one  most  favorable  to  the 
range ;  therefore,  suppose  the  fragments  to  leave  the  wheel  at 
that  angle  and  use  the  formula, 

•z/2 
Horizontal  distance  =  b  =  —  which  transposes  to  v  =  ^/^  g 

RULE. 

Multiply  the  horizontal  distance  by  32.2,  and  the  square  root 
of  the  product  is  the  slowest  possible  rim-speed  the  wheel  could 
have  had  at  the  time  of  the  accident. 

EXAMPLE. 

A  30-foot  fly-wheel  bursts  from  the  stress  due  to  centri- 
fugal force,  and  fragments  were  thrown  a  distance  of  300  feet 
from  the  place  of  accident.  What  was  the  slowest  possible 
speed  the  wheel  could  have  had  at  the  time  the  accident 
occurred  ?  and  what  was  the  corresponding  number  of  revolu- 
tions per  minute? 


MECHANICS.  285 

Solution : 

-v  —  V300  X  32.2  =  V  9660  =  98.3  feet  per  second. 

The  length  of  the  circumference  of  a  30-foot  wheel  is  94.25 
feet,  therefore  the  fly-wheel  was  running  at  a  speed  not  less  than 

98.3 
60   X  94^5  =  62-6  revolutions  per   minute.     This  calculation 

does  not  prove  that  the  wheel  did  not  run  faster  than  62.6 
revolutions  per  minute  when  it  burst;  it  may  have  revolved  a 
great  deal  faster,  as  it  is  not  at  all  sure  that  any  fragments  left 
the  wheel  at  an  angle  of  45  degrees,  but  it  is  certain  that  the 
speed  of  the  wheel  was  not  slower.  Sometimes  it  may  be  pos- 
sible to  settle  upon  the  angle  at  which  a  certain  fragment  left 
the  wheel  by  noticing  traces  and  marks  where  it  went,  and, 
figuring  from  the  angle  and  the  range,  a  pretty  fair  idea  of  the 
bursting  speed  may  be  obtained.  (See  formula  on  page  283). 


Force,  Energy  and  Power. 

Force  is  a  pressure  expressed  in  a  push  or  a  pull. 

Energy  is  the  ability  to  do  work.  It  is  divided  into  poten- 
tial energy  and  kinetic  energy. 

Potential  energy  is  the  ability  of  a  body  to  perform  work 
at  any  time  when  it  is  set  free  to  do  so. 

Kinetic  energy  is  the  ability  of  a  moving  body  to  do  work 
when  its  motion  is  arrested.  Kinetic  energy  is  very  frequently 
called  "  stored-up  energy." 

Work  is  overcoming  resistance  through  space.  In  the 
English  system  of  weights  and  measures  the  common  unit  of 
work  is  the  foot-pound. 

Power  is  the  rate  of  doing  work.  Work  is  an  expression 
entirely  independent  of  time,  but  power  always  takes  time  into 
consideration.  For  instance,  to  lift  one  pound  one  foot  is  one 
foot-pound  of  work,  no  matter  in  what  time  it  is  done,  but  it 
takes  60  times  as  much  power  to  do  it  in  one  second  as  it  would 
lake  to  do  it  in  one  minute. 

Inertia. 

Inertia  is  the  inability  of  dead  bodies  to  change  either  their 
state  of  rest  or  motion.  In  order  to  bring  about  any  change, 
either  of  motion  or  rest,  dead  bodies  must  always  be  acted  upon 
by  some  outside  force. 

Resistance  due  to  inertia  is  the  resistance  which  a  dead 
body  free  to  move  presents  to  any  external  force  acting  tq 
change  either  its  state  of  motion  or  rest. 


286  MECHANICS. 


Mass. 


The  mass  of  a  body  is  the  quantity  of  matter  which  it  con- 
tains. By  common  consent  the  unit  of  mass  is,  in  mechanics, 
considered  to  be  that  quantity  of  matter  to  which  one  unit  of 
force  can  give  one  unit  of  acceleration  in  one  unit  of  time  ; 
therefore,  when  the  weight  of  a  body  is  divided  by  acceleration 
of  gravity,  the  Quotient  is  the  mass  of  the  body.  Thus  : 


W  =  m  X  g 

-W 


Momentum. 

The  product  of  the  mass  of  a  moving  body  and  its  velocity 
is  called  its  momentum  or,  also,  its  quantity  of  motion.  The 
unit  for  momentum  is  the  product  when  unit  of  mass  is  multi- 
plied by  unit  of  velocity  per  second.  In  mechanical  calcula- 
tions, using  English  weights  and  measures,  the  unit  of  mass  is 
weight  divided  by  32.2;  therefore,  unit  of  momentum  will  be: 
Weight  of  the  moving  body  in  pounds  multiplied  by  velocity  in 
feet  per  second  and  the  product  divided  by  32.2.  Thus  : 

q  =  m  X  v  m  =  mass  =  —  ;     therefore, 

g 


q  •=.  Momentum,  or  quantity  of  motion. 
W  —  Weight  of  moving  body  in  pounds. 
V  —  Velocity  of  moving  body  in  feet  per  second. 
g=  Acceleration  of  gravity. 

—  —  is  the  formula  by  which   the  time  in  a  free  fall  is 

obtained,  and,  consequently,  the  momentum  of  a  falling  body 
can  also  be  expressed  by  the  product  of  the  weight  of  the  body 
in  pounds  and  the  time  in  seconds  during  the  fall.  This  product 
is  usually  called  "  time  effect." 

Impulse. 

The  product  of  the  force  and  the  time  in  which  it  is  acting 
as  a  blow  against  a  body  is  called  impulse,  and  it  is  always 
of  the  same  numerical  value  as  the  momentum  of  the  moving 
body. 


MECHANICS.  287 

Kinetic  Energy. 

The  kinetic  energy  stored  in  any  moving  body  is  always 
expressed  in  foot-pounds,  by  the  product  of  the  force  in  pounds 
acting  to  overcome  the  inertia  of  the  body,  and  the  distance  in 
feet  through  which  the  force  was  acting  in  starting  the  body, 
and  is  always  equal  to  the  weight  of  the  body  multiplied  by  the 
square  of  the  velocity  and  this  product  divided  by  twice  the 
acceleration  of  gravity.  Thus : 

K  =  W  X  t/2 

2* 

K  —  Kinetic  energy  in  foot-pounds. 
IV  =  Weight  of  the  body  in  pounds. 
•v  =  Velocity  of  the  body  in  feet  per  second. 
2  g  —  64.4. 

In  a  free  fall  the  height,  /;,  corresponding  to    a  given 

,0 

velocity,  is  found  by  the  formula,  -^- ;  therefore,  K '=  W  X  h. 

Thus,  multiplying  the  weight  of  a  moving  body  by  the  height 
which  in  a  free  fall  corresponds  to  its  velocity,  the  product  will 
be  the  kinetic  energy  stored  in  the  body. 

frTX-z/2 
The  formula  K  =  — ^ transposes  to  K  —  ^  m  v'1. 

Hence  the  simple  rule : 

Multiply  half  the  mass  of  a  moving  body  by  the  square  of 
its  velocity  in  feet  per  second,  and  the  product  is  the  kinetic 
energy  in  loot-pounds  stored  in  the  body. 

The  kinetic  energy  stored  in  any  moving  body  always 
represents  a  corresponding  amount  of  mechanical  work  which 
is  required  in  order  to  again  bring  the  body  to  rest. 

EXAMPLE. 

A  body  weighing  1610  pounds  is  moving  at    a  constant 
velocity  of  18  feet  per  second.      How    many  foot-pounds  of 
kinetic  energy  is  stored  in  the  body  ? 
Solution : 

W  X  7/2       1610  X  18  X  18 
K  =  — 2 =  r    — eO ==  8'100  foot-pounds. 

If  this  moving  body  was  brought  to  rest  and  all  its  stored 
energy  could  be  utilized  to  do  work  it  could  lift  8,100  pounds 
one  foot,  or  it  could  lift  81  pounds  100  feet,  or  any  other  combi- 
nation of  distance  and  resistance  which,  when  multiplied  by 
one  another,  will  give  8,100  foot-pounds. 

It  is  very  important  always  to  keep  in  mind  a  clear  dis- 
tinction between  work  ?CN\  power,  as  power  is  the  rate  of  doing 
work,  and  time  must,  therefore,  always  be  considered  in  the 
question  of  power.  For  instance,  when  33,000  foot-pounds  of 


288  MECHANICS. 

work  is  performed  in  one  minute  it  is  said  to  be  one  horse-power ; 
therefore,  if  this  32,400  foot-pounds  of  energy  was  utilized  to  do 
work  and  used  up  in  one  minute,  it  would  do  work  at  a  rate  of 
fi^lro  —  M  horse-power,  but  if  utilized  during  a  time  of  two 
minutes  it  would  only  do  work  at  a  rate  of  |i  horse-power,  or  if 
utilized  in  a  second  the  rate  of  work  would  be  i*  X  60  =  58  |f 
horse-power,  etc. 

To  Calculate  the  Force  of  a  Blow. 

The  force  of  a  blow  may  be  calculated  by  the  change 
it  produces.  For  instance,  a  drop-hammer  weighing  800  pounds 
drops  three  feet,  and  compresses  the  hot  iron  on  the  anvil  % 
inch.  How  much  is  the  average  force?  (%  inch  =  Vtsfoot). 

The  kinetic  energy  stored  in  the  hammer  at  the  moment  it 
commences  to  compress  the  iron  is  800  X  3  =  2400  foot-pounds. 

The  average  force  =  240°  =  115,200  Ibs. 
Vis 

In  the  above  example,  friction  is  neglected. 

The  shorter  the  duration  of  the  blow  the  more  intense  it 
will  be.  Therefore  the  force  of  the  hammer  mentioned  above,  if, 
instead  of  striking  against  hot  iron,  compressing  it  %  inch,  had 
been  struck  against  cold  iron,  compressing  it  only  a  few  thou- 
sandths, the  blow  would  have  been  as  many  times  more  intense 
as  the  duration  of  the  blow  had  been  shorter.  Therefore  it  is 
entirely  meaningless  to  say  that  a  drop-hammer  or  any  other 
similar  machine  is  giving  a  blow  of  any  certain  number  of 
pounds  by  falling  a  certain  number  of  feet,  because  the  in- 
tensity of  the  blow  will  depend  upon  its  duration. 

Formulas  for  Force,  Acceleration  and  Motion. 

From  the  laws  of  gravitation,  it  is  known  that  when  one 
pound  of  force  acts  upon  one  pound  of  matter  it  produces  an 
acceleration  of  82.2  feet  per  second  each  successive  second  as 
long  as  the  force  continues  to  act. 

From  Newton's  laws  of  motion,  it  is  known  that  the  motion 
is  always  in  proportion  to  the  force  by  which  it  is  produced  ; 
therefore,  when  one  pound  of  force  acts  for  one  second  upon  32.2 
pounds  of  matter,  it  will  produce  an  acceleration  of  one  foot  per 
second. 

Hence  the  following  formulas: 

;//  —  Mass  of  the  moving  body,  which  is  considered  to  be 
weight  divided  by  32.2. 

f=  Constant  force  in  pounds  acting  on  a  body  free  to 
move. 

G  =  Constant  acceleration  in  feet  per  second  due  to  the 
acting  force,  F. 


MECHANICS.  289 

T=  Time  in  seconds  in  which  the  force  F  acts  upon  a 
body  free  to  move. 

i>  —  final  velocity  acquired  by  the  moving  body  in  the 
time  of  T  seconds. 


G  in 

G  =  -?L  T=  —  ^L=.JL.  vm=-FT 

'    T  G  T        m 

-u^L2L  m  =  J—L  F  =  2^L  T-   vm 

m  v  T  F 

When  a  moving  body  is  arrested  the  product  of  the  resist- 
ance and  time  is  equal  to  its  momentum.    Thus: 


z> 

A  = 


#=  Constant  resistance  in  pounds  acting  against  the 
moving  body. 

The  average  velocity  of  the  moving  body  is  half  of  the  final 
velocity,  and  the  space  'passed  over  by  the  moving  body  when 
acquiring  the  given  velocity  is  half  of  the  final  velocity  in  feet 
per  second  multiplied  by  the  time  in  seconds.  Thus  : 


F-*  Sm 


-  o_ 

"  " 


6*  =  Space  in  feet. 

The  work  in  foot-pounds  required  to  overcome  the  inertia  of 
a  given  body  when  brought  from  a  state  of  rest  to  a  given 
velocity  is  equal  to  the  kinetic  energy  stored  in  the  moving 
body.  Thus  : 


K  —SF—  "!  "l'~  —  F>v  T  —  G  m<v  T 
22  2 

K  =  Kinetic  energy  stored  in  the  moving  body. 


290  MECHANICS. 

The  force  required  to  obtain  a  given  velocity  in  a  given 
time,  when  both  resistance  due  to  inertia  and  resistance  due  to 
friction  is  considered,  is  calculated  by  the  formula: 

Force  —  /Velocity  x  Mass^j  +  (weight  X  coefficient  of  friction). 
\    Time  / 

which  may  be  written  : 

Force  =  (  Velocitv    x  Mass^  +  (resistance  due  to  friction). 
V     Time  / 

IMPORTANT.  —  Always  calculate  the  force  required  to  over- 
come the  resistance  due  to  inertia  and  the  force  required  to 
overcome  the  resistance  due  to  friction  separately,  and  add  the 
two  forces  in  order  to  obtain  the  total  force  required. 

It  is  sometimes  assumed  that  adding  so  much  to  the  mass, 
as  gVof  the  product  of  weight  and  coefficient  of  friction,  should 
give"  the  result  in  one  operation  ;  but  such  an  assumption  is 
erroneous,  because  the  correct  value  ior  the  required  force  is  : 


which  cannot  be  transposed  to 


J?=  Required  force  ;  v  =  velocity  ;  T=  time  ;  W=  -weight 
of  moving  body  in  pounds  ;  ^—  acceleration  due  to  gravity,  or 
32.2  ;  f=-  coefficient  of  friction. 

EXAMPLE.  1. 

A  railroad  train  weighing  225,400  pounds  is  started  from 
rest  to  a  velocity  of  50  feet  per  second  ;  the  road  is  straight  and 
level  ;  the  resistance  due  to  friction  is  assumed  to  remain  con- 
stant and  to  be  1000  pounds.  What  average  constant  pull  in 
pounds  must  be  exerted  by  the  locomotive  at  the  draw-bar  in 
order  to  bring  the  train  up  to  this  speed  in  40  seconds  ? 

Solution  :  v 

For  the  inertia, 

velocity  X  mass  __*A  *  225400 

-BST-        L^_M      =8750ibs. 

For  friction  the  force    =  1000 

Total  force,         9750  Ibs. 

NOTE.—  This  constant  force  of  9750  pounds  has  been  acting 
under  a  uniformly  increasing  velocity  from  rest  or  nothing  at 
the  start,  to  50  feet  per  second  at  the  end  of  40  seconds  ;  therefore, 
the  average  velocity  has  been  half  of  the  final  velocity,  or  25  feet 
;t  second.  The  average  work  of  the  locomotive  in  starting  the 


MECHANICS.  291 

train  during  this  40  seconds  was  25  X  0750  =  243,750  foot- 
pounds per  second,  and  the  horse-power  exerted  by  the  locomo- 
tive on  the  draw-bar  in  starting  this  train  was  2~V?;7(ra  —  443.18 
horse-power,  but  the  power  required  to  keep  this  train  in  motion 
at  a  speed  of  50  feet  per  second  on  a  level  road  will  be  only 
*L*p!!  =  90.91  horse-power.  From  this  i  t  may  be  seen  what  an 

immense  power  there  has  to  be  produced  in  order  to  start  heavy 
machinery  in  a  short  time,  in  comparison  to  the  power  requirea 
to  keep  it  going  after  it  is  started. 

EXAMPLE  2. 

How  far  did  the  train  move  before  it  got  up  to  the  required 
speed  of  50  feet  per  second  ? 

Solution  : 

S=ll7:  =  $1X40 

2  2 

EXAMPLE  3. 

Suppose  that  after  the  train  had  acquired  this  speed  of  50 
feet  a  second,  the  locomotive  was  detached  and  that  the  resist- 
ance due  to  friction  continued  to  be  1000  pounds.  How  many 
seconds  would  the  train  be  kept  in  motion  by  its  momentum  on 
a  level  road  ? 

Solution  : 

225400 

T--  v  w         50  X     099 

^  =  —  iooiT1  =  S5°  seconds- 

EXAMPLE  4. 

How  many  foot-pounds  of  kinetic  energy  is  stored  in  this 
train,  which  weighs  225,400  pounds  and  runs  at  a  constant  speed 
of  50  feet  a  second  ? 

K  =  ^!_X-^-  _  5°a  x  12^"  =  8,750,000  foot-pounds. 
2  2 

EXAMPLE  5. 

How  far  will  this  kinetic  energy  drive  the  train  on  a  hori- 
zontal road  if  we  suppose  the  constant  resistance  due  to  friction, 
as  in  Example  3,  to  be  1000  pounds? 

Solution  : 


Distance  =  er&  =  875000°  =  8750  feet. 

resistance  1000 

When  a  body  free  to  move  is  acted  upon  by  a  constant 
force  the  space  passed  over  increases  as  the  square  of  time. 


292 


MECHANICS. 


EXAMPLE  6. 

Under  the  influence  of  a  constant  force  a  body  moves  five 
feet  the  first  second.  How  far  will  it  move  in  eight  seconds, 
friction  not  considered? 

Solution : 

Distance  =  S2  X  5  =  320  feet. 

Centers. 

Center  of  gravity  is  the  point  in  a  body  about  which  all  its 
parts  can  be  balanced.  If  a  body  is  supported  at  its  center  of 
gravity  the  whole  body  will  remain  at  rest  under  the  action  of 
gravity. 

Center  of  gyration  is  a  point  in  a  rotating  body  at  which 
the  whole  mass  could  be  concentrated  (theoretically)  without 
altering  the  resistance,  due  to  the  inertia  of  the  body,  to  angular 
acceleration  or  retardation. 

Center  of  oscillation  is  a  point  at  which,  if  the  whole  matter 
of  a  suspended  body  was  collected,  the  time  of  oscillation 
would  be  the  same  as  it  is  in  the  actual  form  of  the  body. 

Center  of  percussion  is  a  point  in  a  body  moving  about  a 
fixed  axis  at  which  it  may  strike  an  obstacle  without  communi- 
cating the  shock  to  the  axis. 

Moments. 

The  measures  of  tendency  to 
produce  motion  about  a  fixed  point 
or  axis,  is  called  moment.  The  pro- 
duct of  the  length  of  a  lever  and  the 
force  acting  on  the  end  of  it.  tending 
to  swing  it  around  its  center,  is  called 
the  moment  of  force  or  the  statical 
moment,  and  may  be  expressed  in 

either  foot-pounds  or  inch-pounds.  In  Fig.  3,  the  arm  is  18 
inches  long  and  the  force  is  40  pounds;  the  moment  is  18  X  40 
=  720  inch-pounds,  or  ll/2  X  40  =  00  foot-pounds. 


Levers. 

When  a  lever  is  balanced,  the  distance  a, 
multiplied  by  the  weight  w,  is  always  equal  to 
the   distance    b,  multiplied    by  the   force    F. 
In  a  bent  lever  (as  Fig.  5)  it  is  not  the  length 
of  the  lever  but  the  distance  from  the  fulcrum 
at   right  angles  to   the   line  in  which  the  force  is 
acting,  that  must  be  multiplied.     Thus: 
a  X  iv  —  b  X  F. 

In  Fig.  6,  the  force  is  acting  at  a  right  angle 
to  the  lever,  and,  therefore,  the  distance  a  is  equal 
to  the  length  of  the  long  end  of  the  lever. 


FIG.  4. 


MECHANICS.  £93 

The  force  is  applied  to  more  advantage  in          FIG.  e. 
Fig.  6  than  in  Fig.  5.    As  a  rule,  the  force  should 
always  be  applied  so  as  to  act  at  right  angles  to 
the  lever. 

Radius  of  Gyration. 

The  radius  of  gyration  of  a  rotating  body  is  the  distance 
from  its  center  of  rotation  to  its  center  of  gyration. 

moment  oT 


Radius  of  gyration  - 

^rnass  ol  rotating  body 

or,  for  a  plane  surface: 


Radius  of  gyration  =  iyrta 

*    area  of  surface 

The  radius  of  gyration  of  a  round,  solid  disc,  such  as  a  grind- 
stone, when  rotating  on  its  shaft,  is  equal  to  its  geometrical 
radius  multiplied  by  N/^~or  radius  multiplied  by  0.7071  very 
nearly.  The  radius  of  gyration  of  a  round  disc,  if  indefinitely 
thin  and  rotating  about  one  of  its  diameters,  is  equal  to  radius 
divided  by  2.  The  radius  of  gyration  of  a  ring,  of  uniform  cross- 
section,  rotating  about  its  center,  such  as  a  rim  of  a  fly-wheel 
when  rotating  on  its  shaft,  is: 


Radius  of  gyration  =  -y  — 


R  —  Outside  radius. 
r  —  Inside  radius. 

The  radius  of  gyration  of  a  hollow  circle  when  rotating 
about  one  of  its  diameters  is : 


Radius  of  gyration  =  •%/ "*"  r 

'       4 


R  —  Outside  radius. 
r  =  Inside  radius. 

Moment  of  Inertia. 

The  moment  of  inertia  is  a  mathematical  expression  used 
in  mechanical  calculations.  It  is  an  expression  causing  con- 
siderable ambiguity,  as  it  is  not  always  used  to  mean  the  same 
thing. 

The  least  rectangular  moment  of  inertia,  as  used  when 
calculating  transverse  strength  of  beams,  columns,  etc.,  is  the 
sum  of  the  products  of  all  the  elementary  areas  of  cross-sections 
into  the  square  of  their  distances  from  the  axis  of  rotation. 
The  axis  of  rotation  is  considered  to  pass  through  the  center 
of  gravity  of  the  section. 


294 


MECHANICS. 


The  least  rectangular  moment  of  inertia  is  always  equal 
to  the  area  of  surface  of  cross-section,  multiplied  by  the  square 
of  the  radius  of  gyration,  when  the  surface  is  assumed  to  rotate 
about  the  neutral  axis  of  the  section. 

Mathematicians  calculate  the  moment  of  inertia  by  means 
of  the  higher  mathematics,  but  it  may  also  be  calculated 
approximately  by  dividing  the  cross-section  of  the  beam  into 
any  convenient  number  of  small  strips  and  multiplying  the  area 
of  each  strip  by  the  square  of  its  distance  from  its  center-line  to 
the  neutral  axis,  and  the  sum  of  these  products  is  the  moment 
of  inertia,  very  nearly. 

The  narrower  each  strip  is  taken,  the  more  exact  the  result 
will  be  ;  but  it  will  always  be  a  trifle  too  small. 

EXAMPLE  1. 

Find  approximately  the  rectangular  moment  of  inertia  for 
a  surface  (or  section  of  [a  beam)  6"  X  2",  about  its  axis  x  y. 
(See  Fig.  7.) 

Divide  the  surface  into  narrow  strips,  as  #,  #,  c,  d,  e,  f^  g, 
h,  i,j\  k,  I,  and  multiply  each  strip  by  the  square  of  its  distance 
from  the  neutral  axis,  x  y,  and  the  sum  of  these  products  is  the 
moment  of  inertia  of  the  surface. 

---  2^  —  >\         FIG.  7. 


a  -  2 
b  =  2 


X  (2^)2  =  7.5625 
X  (2X)2  =  5.0625 


c  —  2  X  y2  X  (IX)2  =  3.0625 
d=  2  X  ft  X  (IX)2  =  1.5625 
e  =  2  X  ft  X  (  X)2  =  0.5625 
/—  2  X  y2  X  (  X)2  =  0.0625 
^=2  X  #  X  (  X)2  =  0.0625 
// :  =  2  X  #  X  (  X)2  =  0.5625 
2  —  2  X  y2  X  (IX)2  =  1.5625 
y  =  2  X  #  X  (IX)2  =  3.0625 
k  —  2  X  Yz  X  (2X)2  =  5.0625 
1=  2  X  Yz  X  (2X)2  =  7.5625 
Moment  of  inertia  =  35^.75  (approximately). 

The  correct  value   for  the  least  rectangular  moment  of 
inertia  for  such  a  surface  is  obtained  by  the  formula, 


(Depth)3  X  width 


and  for  Fig.  7  will  be 


X  2 


=  36.  Thus,  the 


12  12 

approximate  rule  gives  results  a  trifle  too  small,  but  if  the  sur- 
face had  been  divided  into  smaller  strips,  the  result  would  have 
been  more  correct. 

Radius  of  gyration  for  this  surface,  when  rotating  about 
the  axis  x- y,  is  : 


moment  of  inertia 
area 


=  1.78  inches. 


MECHANICS. 


295 


EXAMPLE  2. 

Find  by  approximation  the  rectangular  moment  of  inertia 
for  a  surface,  as  Fig.  8,  (the  sectional  area  of  an  I  beam)  about 
the  axis  x  y. 

When  the  beam  is  symmetrical,  the  neutral  axis  is  at  an 
equal  distance  from  the  upper  and  lower  side,  and  the  moment 
of  inertia  for  the  upper  and  lower  half  of  the  beam  is  equal  ; 
consequently,  when  calculating  momentof  inertia  for  a  surface 
like  Figs.  8  and  7,  it  is  only  nec- 
essary to  calculate  the  moment 
of  inertia  for  half  the  beam, 
and  multiply  by  2  in  order  to 
get  the  moment  of  the  whole 
beam. 


Solution  : 
a  =  3  X  >/2  X 

/;   =  3  X   %    X 

c  =1  X  Y2  X 

d—  1  X 

e  =  1  X 

/=  1  X 


X  (IX)'2  = 
X  (  K)-2= 
X  (  X)2  = 


=  11.34375 

nr     7.59375 

=  1.53125 
0.78125 
0.28125 
0.03125 


Moment  of  inertia  =  21.5625 
Moment  of  inertia  =  21.5625 


for  upper  half, 
for  lower  half, 
for  beam  (approximately). 


Moment  of  inertia  =  43.125 

Area  of  cross-section  of  beam  is  10  square  inches. 

Radius  of  gyration  =  ^43'125  =  2.07  inches. 

EXAMPLE  3. 

Find  approximately  the  moment  of  inertia  of  a  surface,  as 
Fig.  9  (usual  section  for  cast-iron  beams),  about  the  axis,  x  jr9 
passing  through  the  center  of  gravity  of  the  surface. 

In  shapes  of  this  kind  the  axis  through  the  center  of  gravity 
is  not  at  an  equal  distance  from  the  upper  and  lower  side,  but  it 
can  be  obtained  experimentally  by  cutting  a  templet  to  the 
exact  shape  and  size  of  the  surface  and  balancing  it  over  a 
knife's  edge,  or  it  maybe  calculated  by  the  principle  of  moments, 
as  shown  in  this  example.  Divide  the  surfaces  into  three 
rectangles,  the  upper  flange,  the  web  and  the  lower  flange. 
Assume  some  line  as  the  axis,  for  instance,  the  line  n  m,  which 
is  the  center  line  through  the  lower  flange;  multiply  the  area  of 
each  rectangle  by  the  distance  of  its  center  of  gravity  from  the 
axis  n  in,  and  add  the  products.  Divide  this  sum  by  the  area  of 
the  entire  section,  and  the  quotient  is  the  distance  between  the 
center  of  gravity  of  the  section  and  the  axis  n  m. 


296 


MECHANICS. 


-2— 


FIG.  9. 


SOLVING  FOR  CENTER  OF  GRAVITY  I 

(AREA.)  (DISTANCE.) 

Area  of  upper  flange  =  2X1  =  2  square  inches  X  5  =10 
Area  of  web  —4X1=4  square  inches  X  2)4  =  10 

Area  of  lower  flange    =4X1  =  4  square  inches  X  0      =0 

10  20 

and  20  divided  by  10  =  2''  which  is  the  distance  from  the  center 
of  gravity  of  the  lower  flange  to  center  of  gravity  of  the  section 
of  the  beam,  or  the  neutral  axis  x  y. 

SOLVING  FOR   MOMENT   OF   INERTIA  : 

=  10.56250 
=  7.56250 
=  2.53175 


X   l/2 
X     2 


d  — 

e  = 
f= 


j—  1 
k  =  4 
/—  4 


X  &] 

X 

xxx 

XXX  (IX)2  = 
X  X  (  X)2  = 

X  X  (  X)2  = 

l/2    X    (    X)2  = 

X  x  (ix)2  = 
JA  X 


1.53225 
0.78125 
0.28125 
0.03125 
0.03125 
0.28125 
0.78125 
6.12500 
X  (2#)2  =  10.12500 


Moment  of  inertia  of  beam  =  40.6266  (approximately). 
Area  of  cross-section  of  beam  =  10  square  inches 


Radius  of  gyration  of  beam  =  -——  '-  =  2.015  inches. 


MECHANICS. 


297 


FIG.  1O. 


Polar  Moment  of  Inertia. 

The  polar  moment  of  inertia  is  a  mathematical  expression, 
used  especially  when  calculating  the  torsional  strength  of 
beams,  shafting,  etc.  It  is  very  frequently  denoted  by  the  letter 
/.  The  polar  moment  of  inertia  is  the  sum  of  the  products  of 
each  elementary  area  of  the  sur- 
face multiplied  by  the  square  of 
its  distance  from  the  center  of 
gravity  of  surface.  Suppose  (in 
Fig.  10)  that  the  area  is  divided 
into  circular  rings,  as  a,  b,  c,  d,  e, 
f,g,h,  /,  j\  k,  /,  ;//, «,  o,p,  and  the 
area  of  each  ring  multiplied  by 
the  square  of  its  distance  from 
the  center,  c ;  the  sum  of  all  these 
products  is  the  polar  moment  of 
inertia.  The  moment,  calculated 
this  way,  will  always  be  a  trifle 
too  small,  but  the  smaller  each 
ring  is  taken  the  more  correct  the 
result  will  be.  If  each  ring  could 
be  taken  infinitely  small  the  result  would  be  correct. 

The  polar  moment  of  inertia  is  equal  to  the  square  of  the 
radius  of  gyration  about  the  geometrical  center  of  the  shaft,  mul- 
tiplied by  the  area  of  cross-section  of  the  shaft;  therefore,  for  a 
round,  solid  shaft  (as  the  section  shown  in  Fig.  10),  the  polar 
moment  of  inertia  is  always  expressed  by  the  formula : 
(Radius)4  X  *  (Diameter)4  X  v 

2  '  32 


For  a  hollow,  round  shaft,  the  polar  moment  of  inertia  is 
expressed  by  the  formula, 


D  =  Outside  diameter.  d=  Inside  diameter. 

The  fundamental  principle  for  the  polar  moment  of  inertia 
for  any  shape  of  section  is  that,  if  two  rectangular  moments  of 
inertia  are  taken,  one  being  the  least  rectangular  moment  of 
inertia,  about  an  axis  passing  through  the  center  of  gravity,  and 
the  other,  the  least  rectangular  moment,  about  an  axis  perpendic- 
ular to  the  first  one,  also  through  the  center  of  gravity,  the  sum 
of  those  two  rectangular  moments  is  equal  to  the  polar  moment. 

In  Fig.  10,  the  rectangular  moment  of  inertia  about  the  axis 
x  y  will  be  (diameter)*  X  TT  and  the  rectangular  moment  about 

the  axis  *'  y'  will  also  be  (c*iameter)4  X-T;  thus  the  polar  moment 


will  be   (diameter)4  x  T 


298  MECHANICS. 

EXAMPLE. 

Find  the  polar  moment  of  inertia  and  radius  of  gyration  of  a 
round  shaft  of  4"  diameter. 

Solution  : 


y= 


32 

4«X  3.1416 
32 


Radius  of  gyration  =  ^/  Polar  moment  of  inertia 

area  of  section. 


Radius  of  gyration  =         25-1328 


X  3.1410 
Radius  of  gyration  =  1.414  inches. 

The  term,  moment  of  inertia,  as  used  in  calculating  stored 
energy  in  revolving  bodies,  is  frequently  and  certainly  more 
concisely  called  moment  of  rotation,  and  is  a  mathematical 
expression  by  which  the  effect  of  the  whole  mass  (theoretically) 
is  transferred  to  the  unit  distance  from  center  of  rotation.  This 
term  (moment  of  inertia  or  moment  of  rotation)  is  obtained  by 
multiplying  the  square  of  radius  of  gyration  by  mass  of  moving 
body.*  In  English  measure,  mass  is  taken  as  ^  \  v  of  the  weigh  t 
of  the  revolving  body,  and  the  radius  of  gyration  is  always  taken 
in  feet. 

EXAMPLE. 

A  solid  disc  of  cast-iron,  rotating  about  its  geometrical 
center,  is  six  feet  in  diameter  and  of  such  thickness  that  it  will 
weigh  4073.3  pounds.  What  is  its  moment  of  rotation  or 
moment  of  inertia? 

Radius  of  gyration  =  3  X  \/~f  and  (radius  of  gyration)2  =  32  X  $ 


Mass-  .      126.5 

32.2 

Moment  of  rotation  =  126.5  X  32  X  %  =  569.25. 

NOTE.  —  In  all  such  problems  relating  to  stored  energy  in 
rotating  bodies,  the  radius  of  gyration  is  usually  taken  in  feet 
and  not  in  inches,  as  jn  previous  examples  of  moment  of  inertia, 
when  relating  to  strength  of  material. 

*  Instead  of  multiplying  the  mass  of  the  body  by  the  square  of  radius  of 
gyration  in  feet  and  calling  the  product  moment  of  inertia,  some  writers  multiply  the 
weight  of  the  body  by  the  square  of  the  radius  of  gyration  in  feet  and  call  this 
product  moment  of  inertia.  Thus  last  expression  for  moment  of  inertia,  of  course, 
will  have  a  numerical  value  of  32.2  times  the  first  one.  It  does  not  make  any  differ- 
ence in  the  result  of  the  calculation  whether  weight  or  mass  is  used,  but  the  same 
unit  must  be  adhered  to  throughout  the  whole  calculation. 


MECHANICS.  299 

Angular  Velocity. 

When  a  body  revolves  about  any  axis,  the  parts  furthest 
from  the  axis  of  rotation  move  the  fastest.  The  linear  'velocity 
at  a  radius  of  one  foot  from  the  center  of  rotation  is  called  the 
angular  velocity  of  the  body.  It  is  usually  reckoned  in  feet  per 
second.  The  angular  velocity  of  any  revolving  body  is  ex- 
pressed by  the  formula, 

Fa  =  2  TT  n 

V*  =  Angular  velocity  in  feet  per  second. 
n  —  Number  of  revolutions  per  second. 

RULE. 

Multiply  the  number  of  revolutions  per  second  by  6.2832, 
and  the  product  is  the  angular  velocity  in  feet  per  second. 

EXAMPLE. 

What  is  the  angular  velocity  of  a  fly-wheel  making  300 
revolutions  per  minute? 

Solution  : 

300  revolutions  per  minute  =  5  revolutions  per  second, 
therefore,  angular  velocity  =  6.2832  X  5  =  31.416  feet  per 
second.  Angular  velocity  expresses  the  velocity  at  unit  dis- 
tance from  center  of  rotation  and  in  English  measure  this  unit  is 
feet.  As  already  stated,  the  moment  of  rotation  is  an  expres- 
sion for  the  mass  of  the  rotating  body  (theoretically)  transferred 
to  unit  distance  from  center  of  rotation  ;  the  product  of  angular 
velocity  and  moment  of  rotation  will,  therefore,  be  the 
momentum  of  the  rotating  body.  The  constant  resistance 
which  has  to  be  exerted  at  unit  radius  in  order  to  bring  the 
body  to  rest  in  jT  seconds  will  be  : 


The  resistance  which  has  to  be  exerted  at  any  radius  of  r 
feet  to  bring  the  body  to  rest  in  T  seconds  will  be  : 


R  —  Resistance  in  pounds. 
Fa  —  Angular  velocity  in  feet  per  second. 

/  =  Moment  of  rotation  (also  called  moment  of  inertia). 
The  constant  force  which  has  to  be  exerted  at  unit  radius 
in  order  to  bring  the  body  from  a  state  of  rest  to  an  angular 
velocity  Fa  in  T  seconds  will  be: 


300  MECHANICS. 

The  constant  force  which  has  to  be  exerted  at  any  radius, 
r,  in  order  to  bring  the  body  from  a  state  of  rest  to  an  angular 
velocity  Fa  in  T  seconds  will  be  : 


R  =  Constant  resistance  in  pounds. 

F  =  Constant  force  in  pounds. 

Fa  =  Angular  velocity  in  feet  per  second. 

/  =  Moment  of  rotation  (also  called  moment  of  inertia). 

r  =  Radius  in  feet  at  which  the  force  is  applied. 

T  ~  Time  in  seconds  that  the  force  is  acting. 

EXAMPLE. 

A  fly-wheel  making  120  revolutions  per  minute  and  weigh- 
ing 483  pounds,  is  brought  to  rest  in  two  seconds  by  a  resistance 
acting  at  a  six-inch  radius.  The  radius  of  gyration  of  the  fly- 
wheel is  1.2  feet.  What  is  the  average  force  exerted  against  the 
resistance  during  these  two  seconds  ? 

Solution  : 
120  revolutions  per  minute  =  2  revolutions  per  second. 

Angular  velocity  =  6.2832  X  2  =  12.5664  feet  per  second. 

Moment  of  rotation  =  1.2  X  1.2  X    483-  =21.6 

32.2 

Radius  of  resistance,  6  inches  =  0.5  feet. 

*  =  .12.6664  X  21.6   =  271.43  pounds. 
2  X  0.5 

If  a  rotating  body  is  not  brought  to  rest,  but  only  reduced 
in  speed  to  an  angular  velocity  of  Fa  ,  in  T  seconds,  then  the 
average  force  or  resistance  acting  at  unit  radius  is  : 


F  = 


_   (Fa   —   Fai)7 


The  average  force  which  has  to  be  exerted  at  any  radius  at 
r  feet  to  reduce  the  angular  velocity  to  Fa ,  in  /"seconds  will  be : 


F  = 


Fai)/ 


T  r 
EXAMPLE. 

A  fly-wheel  on  a  punching  machine  weighs  644  pounds,  its 
radius  of  gyration  is  1^  feet,  and  it  makes  at  normal  speed  300 
revolutions  per  minute,  but  when  the  machine  .is  punching  the 


MECHANICS.  301 

speed  is  in  £  of  a  second  reduced  to  a  rate  of  280  revolutions 
per  minute.  What  average  force  has  the  fly-wheel  communi- 
cated to  the  pitch-line  of  a  6-inch  gear  on  the  fly-wheel  shaft  ? 

Solution  : 

The  mass  of  the  fly-wheel  =  644    =  20 

32.2 


The  moment  of  rotation  =  (1J^)2  X  20  =  45 
300  revolutions  per  minute  =  5  revolutions  per  second. 

Angular  velocity  =  5  X  6.2832  =  31.416 
280  revolutions  per  minute  =  4%  revolutions  per  second. 
Corresponding  angular  velocity  =  4^  X  6.2832  =  29.3216 
6-inch  diameter  of  gear  =  3-inch  radius  =  %  foot. 
F  _  (31.416  —  29.3216)  X  45 

ix  X 

F  =  2.0944  X  45  X  5  X  4  =  1884.96  pounds. 

The  kinetic  energy  in  foot-pounds  stored  in  the  revolving 
body  may  be  obtained  by  the  formula  : 

Fa2  X  —  =  kinetic  energy. 
2 

Decreasing    the    angular    velocity    to     F»i,   the   stored-up 
energy  will  also  decrease  to 

W  X  -£. 

and  the  work  done  by  the  revolving  body  will  be 


(Fa2-   Fai2)X    JL 


EXAMPLE   1. 


The  moment  of  rotation  in  a  fly-wheel  is  1040  ;  its  angular 
velocity  is  5  feet  per  second.  What  is  the  stored-up  energy  in 
the  wheel  ? 

Solution : 

Kinetic  energy  =  52  X  !°-*°-  =  13,000  foot-pounds. 

EXAMPLE  2. 

At  certain  intervals,  when  machinery  is  started,  the  angular 
velocity  of  this  fly-wheel  is  reduced  to  ^/2  feet  per  second. 
How  many  foot-pounds  of  energy  has  the  fly-wheel  given  up  in 
helping  to  drive  the  machinery  ? 


3O2  MECHANICS. 

Solution : 

x  =  (52  —  (4>^)2)  X  520 
x  —  (25  —  20#)  X  520 

:r  =  4%   X   520  =  2470  foot-pounds  of  energy  given 
out  by  the  fly-wheel  during  this  change  of  speed. 

EXAMPLE  3. 

How  much  stored  energy  is  left  in  the  wheel  after  its  angu- 
lar velocity  is  reduced  to  4>£  feet  per  second  ? 

Solution : 
K=  (Fa)2       - 


X  520  =  20  #  X  520  =  10,530  foot-pounds. 
The  same  result  may  be  obtained  by  subtracting,  thus : 
13,000  —  2470  =  10,530  foot-pounds. 

Centrifugal  Force. 

The  centrifugal  force  is  the  force  with  which  a  revolving 
body  tends  to  depart  from  its  center  of  motion  and  fly  in  a 
direction  tangent  to  the  path  which  it  describes.  The  centrip- 
etal force  is  the  force  by  which  a  revolving  body  is  prevented 
from  departing  from  the  center  of  motion.  When  the  centri- 
fugal force  exceeds  the  centripetal  force  the  body  will  move 
away  from  the  center  of  motion,  but  if  the  centripetal  force  ex- 
ceeds the  centrifugal  force,  the  body  will  move  toward  the  center 
of  motion.  The  centrifugal  force  in  any  revolving  body  is  equal 
to  the  mass  of  the  body  (see  page  286)  multiplied  by  the  square 
of  its  velocity,  and  this  product  divided  by  the  radius  of  the 
revolving  body. 

W  X  v2      m  X  v2 


~  32.2  X  r  ~        r 
cf—  Centrifugal  force  in  pounds. 

r  =  Radius  in  feet. 

-v  =  Velocity  in  feet  per  second. 
W  =:  Weight  of  moving  body  in  pounds. 
m  =  Mass  of  moving  body.  F|G-  1 1 

EXAMPLE. 

The  weight  #,  in  Fig.  11,  is  four  pounds,  and  the  length  of 
the  string  is  two  feet;  the  weight  is  made  to  swing  around  the 
center  c,  three  revolutions  per  second.  What  is  the  stress  on 
the  string  due  to  centrifugal  force? 


MECHANICS.  303 


Solution : 


The  distance  from  c,  to  the  center  of  the  ball  is  two  feet, 
and  making  three  revolutions  per  second,  the  velocity  will  be 
2  X  3  X  3.1410  X  2  —  37.7  feet  per  second. 

cf=  4  X  37'7  X  37'7  =  88.2  pounds. 
32.2  X  2 

In  metric  measure, 

^      9.81  X  r 

cf=  Centrifugal  force  in  kilograms. 

r  =  Radius  in  meters. 

v  =  Velocity  in  meters  per  second. 
W=  Weight  of  moving  body  in  kilograms. 

EXAMPLE. 

Suppose  that  the  weight  a,  in  Fig.  11,  is  five  kilograms, 
swinging  around  the  center,  c,  one  revolution  per  second  ;  the 
distance  from  a  to  c  isl}4  meters.  What  is  the  stress  on  the 
string  due  to  centrifugal  force  ? 

Solution : 

The  velocity  will  be  1.5  X  3.1416  X  2  —  9.4248  meters  per 
second. 

5  X  9.4248  X  9.4248  =  ^     kilograms. 
9.81  X  iy2 

Friction. 

The  resistance  which  a  body  meets  with  from  the  surface 
on  which  it  moves  is  called  friction.  It  is  called  sliding  friction 
when  one  body  slides  on  another ;  for  instance,  a  sleigh  is  pulled 
along  on  ice — the  friction  between  the  runners  of  the  sleigh  and 
the  ice  is  sliding  friction.  It  is  said  to  be  rolling  friction  when 
one  body  is  rolling  on  another  so  that  new  surfaces  continually 
are  coming  into  contact;  for  instance,  when  a  wagon  is  pulled 
along  a  road,  the  friction  between  the  wheels  and  the  road  is  roll- 
ing friction,  but  the  friction  between  the  wheels  and  their  axles 
is  sliding  friction.  Sliding  friction  varies  greatly  between 
different  materials,  as  everybody  knows  from  daily  observation. 
For  instance,  a  sleigh  with  iron  runners  can  be  pulled  with  less 
effort  on  ice  than  on  sand,  even  if  the  road  is  ever  so  smooth. 
This  is  because  the  friction  between  iron  and  ice  is  a  great 
deal  less  than  the  friction  between  iron  and  sand, 


304  MECHANICS. 

Coefficient  of  Friction. 

The  ratio  between  the  force  required  to  overcome  the  re- 
sistance due  to  friction  and  the  weight  of  a  body  sliding  along 
a  horizontal  plane  is  called  coefficient  of  friction. 

For  instance,  in  Fig.  12  a 
piece  of  iron  weighing  300  Ibs. 
rests  on  a  horizontal  plate  b.  A 
string  fastened  to  «,  goes  over  a 
pulley,  c.  At  the  end  of  the 
string  is  applied  a  weight,  d.  If 
this  weight  is  increased  until 
the  body  a  just  starts  to  move 
along  on  b,  and  the  weight  is  found  to  be  50  pounds,  the  co- 
efficient of  friction  will  be  -50  —  —  =  0.166 
300  6 

When  the  weight  of  a  moving  body  is  multiplied  by  the 
coefficient  of  friction,  the  product  is  the  force  required  to  keep 
the  body  in  motion.  Of  course,  any  pressure  applied  to  the 
moving  body,  perpendicular  to  its  line  of  motion,  may  be  sub- 
stituted for  its  weight.  For  instance,  the  frictional  resistance 
of  the  slide  in  a  slide-valve  engine  is  not  due  to  the  weight  of  the 
valve,  but  to  the  unbalanced  steam  pressure  on  the  valve.  In  all 
cases  the  rule  is : 

Multiply  the  coefficient  of  friction  by  the  pressure  perpen- 
dicular to  the  line  of  motion,  and  the  product  is  the  force 
required  to  overcome  the  frictional  resistance. 

EXAMPLE. 

The  coefficient  of  friction  is  0.1,  and  the  weight  of  the 
sliding  body  is  800  pounds.  What  force  is  required  to  slide  it 
along  a  horizontal  surface  ? 

Solution : 

Force  =  800  X  0.1  =  80  pounds.    ' 

Rolling  Friction. 

If  the  body,  a,  (see  Fig.  12)  was  lifted  up  from  the  plane,  b, 
high  enough  so  that  two  rollers  could  be  placed  between  a  and 
£,  it  would  be  found  that  the  body  would  move  with  much  less 
force  than  50  pounds  because,  instead  of  sliding  friction,  as  in 
the  first  experiment,  it  would  be  rolling  friction.  Suppose  it  is 
found  that  a  commenced  to  move  when  the  load,  d,  was  four 
pounds,  then  the  coefficient  of  friction  for  this  particular  case 

would  be    4-  —  —  —  0.0133 
300       75 

In  these  experiments  the  whole  force  at  d  is  not  used  to 
move  the  load  a,  as  a  small  part  of  it  is  used  to  move  the  pulley 
at  c,  but  in  order  to  make  the  principle  plain,  this  loss  has  not 
been  considered. 


MFCMANICS.  _  305 

Axle  Friction. 

The  friction  between  bearings  and  shafts  is  frequently 
called  axle  friction.  This,  of  course,  is  sliding  friction,  but 
owing  to  the  fact  that  the  surfaces  in  question  are  usually  very 
smooth  and  well  lubricated,  the  coefficient  of  friction  is  smaller 
than  for  ordinary  slides. 

EXAMPLE  2. 

A  fly-wheel  weighs  24,000  pounds,  the  diameter  of  the  shaft 
is  10  inches,  and  the  coefficient  of  friction  in  the  bearings  is  0.08. 
What  force  must  be  applied  20  inches  from  the  center  in  order  to 
keep  the  wheel  turning  ? 

Resistance  due  to  friction  =  24000  X  0.08  =  1920  pounds. 

This  resistance  is  acting  at  a  radius  of  5  inches,  but  the 
force  is  acting  at  a  radius  of  20  inches ;  therefore,  the  required 
force  necessary  to  overcome  friction  will  be  1-2()-*J?  —  480  pounds. 

How  much  power  is  absorbed  by  this  frictional  resistance  if 
the  wheel  is  moving  72  revolutions  per  minute  ? 

Solution : 


72  X  20  X  2  X  3.1416 
12 

=  753.984  feet,  and  753.934  X  480  —  361,912.32  foot-pounds  and 


The  space  moved  through  by  the  force  is 


Horse=Power  Absorbed  by  Friction  in  Bearings. 

The  horse-power  absorbed  by  the  friction  in  the  bearings 
for  any  shaft  may  be  figured  directly  by  the  formula, 

HP—  W/x/x  n  X  3.1416  X  d 

S30"00"X  12 
This  reduces  to : 

H-P  =WXfXnXdX  0.000008 
H-P  —  Horse-power  absorbed  by  friction. 
W  =  Load  on  bearings  in  pounds. 
d=  Diameter  of  shaft  in  inches. 
/•=.  Coefficient  of  friction. 
n  =  Number  of  revolutions  per  minute. 

Calculating  the  previous  example  by  this  formula,  we  have: 

H-P  =  24000  X  0.08  X  72  X  10  X  0.000008  =  11. 06  horse- 
power, which  is  practically  the  same  as  figured  before. 


3o6 


MECHANICS. 


Angle  of  Friction. 

Suppose,  instead  of  using  the  string  and  the  weight  d  (see 
Fig.  12),  that  one  end  of  the  plane  is  lifted  until  a  commences  to 
slide ;  the  angle  between  b  and  the  horizontal  line,  when  a  com- 
mences to  move,  is  called  the  angle  of  friction.  The  coefficient 
of  friction  may  also  be  calculated  from  the  angle  of  friction,  thus  : 
If  the  body  commences  to  slide  under  an  angle  of  a  degrees,  the 
coefficient  of  friction  will  be  sm-  a  —  tang.  a.  Thus,  the  coefficient 

cos.  a 

of  friction  is  always  equal  to  tangent  of  the  angle  of  friction. 

Rules  for  Friction. 

1.  Friction  is  in  direct  proportion   to   the   pressure  with 
which  the  bodies  are  bearing  against  each  other. 

2.  Friction  is  dependent  upon  the  qualities  of  the  surfaces 
of  contact. 

3.  The  velocity  has,  within  ordinary  limits,  no  influence  on 
the  value  of  the  coefficient  of  friction. 

4.  Sliding  friction  is  greater  than  rolling  friction. 

5.  Friction  offers  greater  resistance  against  starting  a  body 
than  it  does  after  it  is  set  in  motion. 

6.  The  area  of  surfaces  of  contact  has,  within  ordinary 
limits,  no  influence  upon  the  value  of  the  coefficient  of  friction, 
but  if  they  are  unproportionally  large  or  small  the  friction  will 
increase. 


TABLE  No.  36.— Coefficient  of  Friction. 


MATERIALS. 

SLIDES. 

BEARINGS. 

Well 
Lubri- 
cated. 

Not  well 
Lubri- 
cated. 

Well 
Lubri- 
cated. 

Not  well 
Lubri- 
cated. 

Cast-iron  on  wrought  iron  .... 
Cast-iron  on  cast-iron  .  .  ... 

0.08 
0.08 
0.08 
0.10 

0.16 
0.16 
0.20 
0.20 

0.05 
0.05 
0.05 
0.05 

0.075 
0.075 
0.075 
0.075 

Wrought  iron  on  brass  
Wrought  iron  on  wrought  iron  .  . 

Friction  in  Machinery. 

When  the  surfaces  are  good  the  frictional  resistance  for 
slides  may  be  assumed  as  10  per  cent.,  more  or  less,  according 
to  the  conditions  of  the  surfaces.  It  is  always  well  not  to  take 
the  coefficient  of  friction  too  small;  it  is  better  to  be  on  the 
safe  side  and  allow  power  enough  for  friction.  In  bearings  for 
machinery,  the  frictional  resistance  ought  not  to  absorb  over 
six  per  cent.  If  more  is  wasted  in  friction,  there  is  a  chance 
for  improvement. 


MECHANICS. 


3°7 


FIG.  13. 


150  Ibs. 


Pulley  Blocks. 

When  friction  is  not  considered,  the 
force  and  the  load  will  be  equal  in  a  single 
fixed  pulley  (as  A,  Fig.  13). 

Thus,. a  single  fixed  pulley  does  not 
accomplish  anything  further  than  to  change 
the  direction  of  motion.  In  a  single  movable 
pulley  (as  -at  B,  Fig.  13),  the  force  is  equal 
to  only  half  the  load;  thus,  75  pounds  of 
force  will  lift  150  pounds  of  load,  but  the 
force  must  act  through  twice  the  space  that 
the  load  is  moved.  The  tension  in  any  part 
of  the  rope  in  B  is  half  of  the  load  W\  thus,  when  the  load  is 
150  pounds  the  tension  in  the  rope  is  75  pounds,  when  arranged 
at  B,  but  it  is  150  pounds  when  arranged!  as  at  A. 

Fig.  14  shows  a  pair  of  single  sheave  pulley  blocks  in 
position  to  pull  a  car;  when  the  blocks  are  arranged  as  at  A, 
and  friction  is  not  consid- 
ered, a  force  of  100  pounds 
on  the  hauling  part  of  the 
rope  exerts  a  force  of  300 
pounds  on  the  post,  but  only 
200  pounds  on  the  car;  but, 
turning  the  blocks  end  for 
end,  as  shown  at  B,  a  force 
of  100  pounds  on  the  hauling 
part  of  the  rope  exerts  a 
force  of  300  pounds  on  the  car  and  200  pounds  on  the  post. 
This  is  a  point  well  worth  remembering  when  using  pulley  blocks. 
Suppose,  for  instance,  that  a  man  exerted  a  force  of  100 
pounds  on  the  hauling  part,  and  that  it  required  250  pounds 
of  force  to  move  the  car ;  if  he  used  the  pulley  blocks  as 
shown  at  A,  his  work  would  be  useless,  as  far  as  moving  the  car 
is  concerned,  as  he  could  not  do  it,  but  turning  his  blocks  end 
for  end  he  could  accomplish  the  desired  result.  Always  re- 
member whenever  it  is  possible  to  have  the  hauling  part  of  the 
rope  coming  from  the  movable  block  and  pull  in  the  same 
direction  as  the  load  is  moving. 


FIG.    14. 


Friction  in  Pulley  Blocks. 

In  practical  work,  friction  will  have  some  influence,  and,  to 
a  certain  extent,  change  these  results,  because  some  of  the  ten- 
sion in  the  rope  is  lost  by  friction  in  each  sheave  the  rope  passes 
over,  therefore  the  tension  in  each  following  part  of  the  rope  is 
always  less  than  it  was  in  the  preceding  part.  This  loss  must 
be  obtained  from  experiments.  In  good  pulley  blocks,  having 
roller  bearings,  this  loss  is  probably  not  more  than  0.1,  and  we 


MECHANICS. 

get  a  useful  effect  of  0.9  of  the  force  from  one  part  of  the  rope 
to  the  next  ;  therefore,  when  friction  is  considered,  the  useful 
effect  in  the  following  cases  will  be  : 

In  single  sheave  blocks  having  the  hauling  part  from  the 
movable  block  (pulling  with  the  load  as  in  B,  Fig.  14). 

W=F(\  +  0.9  +  0.92) 
W=  F  X  2.71 

In  single  sheave  blocks  having  the  hauling  part  from  the 
fixed  block  (pulling  against  the  load  as  in  A,  Fig.  14), 

W  =  F  (0.9  +  0.92) 
tV=FXl.1\ 

In  double  sheave  blocks  having  the  hauling  part  from  the 
movable  block, 


0.92  +  0.93  +  0.94) 
W=FX  4.1 

In  double  sheave  blocks  having  the  hauling  part  from  the 
fixed  block, 

W  =  F  (0.9  +  0.92  +  0.93  +  0.94) 
IV=FX  3.1 


Differential  Pulley  Blocks. 

In  a  differential  pulley  block  (see  Fig.  15),  the 
proportion  between  the  force  and  the  weight, 
when  friction  is  neglected,  is  expressed  by  the 
formula : 

F_  W  X  (R  —  r) 
2X7? 


The  actual  force  required  to  lift  a  weight  by 
such  a  pulley  block  is  about  three  times  the 
theoretical  force,  as  calculated  above. 


Inclined  Plane. 

When  a  weight  is  pulled  upward  FIG.  16. 

on  an  inclined  plane,  as  shown  in  Fig. 
16,  and  the  force  F  is  acting  parallel 
to  the  plane,  the  required  force  for 
moving  the  body  will  be  F  —  W  X 
sin.  a  plus  friction,  and  the  perpen-  *' 
dicular  pressure  P.  against  the  plane 
will  be  IV  X  cos,  «.  r_- eos.-a- 


FIG.  15. 


MECHANICS.  309 


EXAMPLE  1. 


The  weight,  W,  (Fig.  16)  is  100  pounds ;  the  angle  a  is  30°. 
What  force,  F,  is  required  to  sustain  this  weight,  friction  not 
considered  ? 

Solution  : 

Sin.  30°  =  0.5 

Thus: 

F  —  W  X  sin.  30°  =  100  X  0.5  =  50  pounds. 

EXAMPLE  2. 

What  is  the  perpendicular  pressure  under  conditions  stated 
in  Example  1  ? 

Solution : 

P  =  IV  X  cos.  a  =  100  X  0.86603  =  86.6  pounds. 

Therefore,  the  frictional  resistance  between  the  sliding 
body  and  the  inclined  plane  will  be  only  what  is  due  to  86.6 
pounds  pressure ;  in  other  words,  the  force  required  to  over- 
come friction  will  be  W  X  f  X  cos.  a. 

EXAMPLE  3. 

What  force  is  required  to  move  the  body  mentioned  in 
Example  1  when  friction  is  also  considered,  taking  coefficient 
of  friction,  F,  as  0.15? 

Solution : 

F  =  W  (sin.  a  +  cos.  a  X  /) 
.F  =  100X  (0. 5  -f- 0.86603X0. 15)=  100  X  0.6290  =62. 99 pounds. 

NOTE. — This  is  the  force  required  for  moving  the  load. 
In  order  to  put  it  in  motion  more  force  must  be  applied,  varying 
according  to  velocity,  but  after  motion  is  commenced  the  speed 
would  be,  under  these  conditions,  maintained  forever  by  this 
force  of  62.99  pounds. 

When  a  load  is  moving  down  an  inclined  plane  the  force 
due  to  W  X  sin.  a  will  assist  in  moving  the  body,  and  if  the 
product  W  X  sin.  a  exceeds  the  product  W  X  cos.  a  X  f  the 
body  will  slide  by  itself.  For  instance,  in  the  body  mentioned 
in  the  previous  example,  the  force  required  to  overcome  gravity, 
regardless  of  friction,  is  50  pounds,  and  the  force  required  to 
overcome  friction  is  12.99  pounds;  thus,  if  the  body  should  be 
let  down  the  plane  instead  of  pulled  up,  it  would  have  to  be 
held  back  with  a  force  of  50  —  12.99  =  37.01  pounds. 

NOTE. — When  the  incline  is  less  than  1  in  35,  cosine  is  so 
nearly  equal  to  1  that  it  may  be  neglected,  and  the  force  required 
to  overcome  friction  may  be  considered  to  be  the  same  as  on  a 
level  plane.  For  instance,  a  horse  is  pulling  a  load  and  ascend- 
ing a  gradient  of  1  in  35 ;  if  the  tractive  force  required  to  pull 
the  load  on  a  level  road  was  30  pounds  and  the  weight  of  the 
load  was  1400  pounds,  when  ascending  the  hill,  the  horse  will  first 


310  MECHANICS. 

have  to  exert  a  force  of  30  pounds,  which  is  all  due  to  friction, 
but  beside  that  he  must  also  exert  a  force  of  ^  times  1400  —  40 
pounds ;  thus  the  total  pull  exerted  by  the  horse  will  be  70 
pounds. 

Inclined  Plane  With  the  Force  Acting  Parallel  to  the  Base. 

When  the  pressure  is  continually  FIG.  17. 

acting  in  a  line  parallel  to  the  base  of  ./       F" 

the  incline,  as  F,  (see  Fig.  17)  which  J/?£^/'^  * 

is  frequently  the  case  in  mechanical      F^*  ^2^^  ** 

movements,  as  for  instance,  in  screws,  /*          •  *f 

some  kinds  of  cam  motions,  etc.,  it      /^      ^ ±_ 

will  require  more  force  to  move  the 

body  than  it  would  if  the  force  was      i* Cos— a >i 

acting  parallel  to  the  incline.     When 

force  acts  parallel  to  the  base,  as  in  Fig.  17,  the  force  required 

to  move  the  body,  if  friction  is  not  considered,  will  be : 

F  —  WK  sm'^  =  W  X  tang,  a 
cos.  a 

EXAMPLE  1. 

What  force  is  required    to  move  100  pounds  upward  an 
incline  of  30°,  as  in  Example  1,  excepting  that  the  force  is  acting 
parallel  to  the  base  instead  of  parallel  to  the  incline  ? 
Solution : 

F—WY.  tang.  30° 

F—  W  X  100  X  0.57735  =  57.74  pounds. 

When  both  the  friction  and  the  weight  of  the  body  are  con- 
sidered, the  force  required  to  move  the  body  will  be  : 

/r_  W  x     s'm.a  +  (f  X  cos. a) 


—  (fX  sin.rt) 
EXAMPLE  2. 

What  force  is  required  to  move  100  pounds  upward  an  in- 
cline of  30°  (as  in  Example  1)  if  the  force  is  acting  parallel  to 
the  base  line  instead  of  parallel  to  the  incline ;    coefficient  of 
friction  is  supposed  to  be  0.15? 
Solution : 

F_  100       sin. 30°  +  (0.15  X  cos. 30°) 
cos.  30°  —  (0.15  X  sin.  30°)~ 
p  _  100  x  0.5  +  (0.15  X  0.86603) 
0.86603  —  (0.15  X  0.5) 

yr^pox  °'5  +  0.1277045 
0.86603  —  0.075 

.F—  100  X  0.7936  —  79.36  pounds. 


MECHANICS. 


311 


NOTE. — From  these  calculations  it  is  seen  that  it  is  more 
advantageous  to  apply  the  force  parallel  to  the  incline  than 
parallel  to  the  base.  When  force  is  applied  parallel  to  the  in- 
cline : 

The  force  required  to  overcome  gravity  —  50  pounds. 

The  force  required  to  overcome  friction  =  12.99  pounds. 
Total  force  —  62.99  pounds. 

When  the  force  is  acting  parallel  to  the  base  : 

The  force  required  to  overcome  gravity  =  57.74  pounds. 

The  force  required  to  overcome  friction  =  21.62  pounds. 
Total  force  =  79.36  pounds. 

Screws. 

When  friction  is  not  considered,  the  force  which  may  be 
exerted  by  a  screw  (see  Fig.  18)  will  be : 

~~ P  R  X  2?r 

W  —  Weight  of  the  load  lifted,  or  force  exerted,  if  the 

screw  acts  as  a  press. 
F  =  Acting  force. 

R  =  Radius  in  inches  at  which  the  force  acts. 
P  =  Pitch  of  screw  in  inches. 

FIG.  18 

i 

Regarding  friction  in 
screws,  the  thread  of  a  screw 
may  be  considered  as  an  in- 
clined plane,  of  which  the 
cos.  is  the  middle  circum- 
ference of  the  screw,  the 
sin.  is  the  pitch,  and  the 
force  is  acting  parallel  to 
the  base.  Hence  the  fol- 
lowing formula : 


F  —  Force,  acting  at  a  radius  of  R  inches. 

W  —  Weight. 

P  =  Pitch  of  screw  in  inches. 

/  =  Coefficient  of  friction,  usually  taken  as  0.15. 

R  =  Radius  in  inches  at  which  the  force  is  acting. 

r  =  Middle  radius  of  screw  in  inches. 

d  =  Middle  diameter  of  screw  in  inches. 


MECHANICS. 

EXAMPLE. 

Find  the  force  required  to  act  on  a  lever  30  inches  long  (see 
Fig.  18)  in  order  to  lift  the  load  W,  which  is  8000  pounds  ?    The 
screw  is  ^-inch  pitch  and  IX'inch  middle  radius  ;    coefficient  of 
friction,  0.15. 
Solution  : 

F  =  8000  X     °-5  +  °-15  x  3-1416  X  2.5        1.25 
2.5  X  3.1416  —  0.15  X  0.5    >      30 


.F=8000  X  X   0.0416  ==  89.6  pounds. 

<  .7  lo 

When  the  screw  has  V  thread,  the  frictional  resistance  will 
be  increased  as  -*-  of  the  angle  a  (see  Fig.  18),  or  equal  to 
secant  of  half  the  angle  of  the  thread.  For  United  States 
standard  screws  the  angle  of  thread  is  60°,  half  the  angle  is  30°, 
and  secant  of  30°  is  1.1547,  and  the  formula  will,  for  United 
States  standard  thread,  become  : 


- 
dK  —  1.15//>         R 

All  the  letters  having  the  same  meaning  as  in  the  formulas 
for  the  square-threaded  screws. 

The  following  table  is  calculated  for  square-threaded  screws, 
the  pitch  of  the  screw  being  double  that  of  the  United  States 
standard  screw  of  same  diameter.  The  depth  of  the  thread  is 
equal  to  its  width.  We  see  no  good  reason  why  the  depth  of  a 
square-threaded  screw  should  be,  as  frequently  given  in  tech- 
nical books,  £§  of  the  pitch  of  the  screw  ;  f  J,  as  given  in  pre- 
vious tables,  is  more  convenient,  and  also  gives  a  little  more 
wearing  surface  to  the  thread.  The  use  of  this  table  is  so 
plain  that  it  needs  very  little  explanation.  In  the  fourth  column 
is  the  area  of  the  outside  diameter  of  the  screw.  In  the  fifth 
column,  the  sectional  area  of  the  screw  at  the  bottom  of  the 
thread,  which  may  be  used  in  calculating  the  tensile  and  crush- 
ing strength  of  the  screw.  Subtracting  the  fifth  column 
from  the  fourth  gives  the  sixth  column,  which  is  the  projected 
area  of  one  thread  ;  this  may  be  used  in  calculating  the  allow- 
able pressure  on  the  thread,  etc.  The  fourteenth  column  gives 
the  tangential  force  which  is  required  to  act  with  a  leverage  of 
one  foot  in  order  to  lift  one  pound  by  the  screw  if  there  was  no 
friction.  The  fifteenth  column  gives  the  total  tangential  force 
required  per  pound  of  load  when  both  load  and  friction  are 
included.  The  sixteenth  column  gives  the  difference  between 
the  fourteenth  and  the  fifteenth  columns,  and  is  the  tangential 
force  absorbed  by  friction  alone.  The  coefficient  of  friction  in 
both  columns  is  assumed  as  0.16.  The  last  four  columns  in 
the  table  give  the  load  or  axial  pressure  which  may  be  allowed 
on  the  screw  corresponding  to  200,  400,  600  and  1000  pounds 
pressure  per  square  inch  of  projected  area  of  screw  thread 
when  the  length  of  the  nut  is  twice  the  diameter  of  the  screw. 


JO 

B3JB  pajoafoad  jo  qoui 

aiEnbs  jad  s'punod  0001°' 

Suipuodsajjoo  aanssajd  jBixy 


«c;£Si 

TH  TH  TH  <N  !M 


r~co 

(M  <N 

I<N 
t~ 


10 

E3JB   pajoafcud  p  ipui 

aienbs   jad    spunod  QOQ   oj 

3uipuods3JJOD  ajnssajd  jBix 


"*  o  in  t—  • 

rH  00  TH  t- 
CO  t—  i— i  i—  ' 
<N  CO  lO  CD 


|0 

E3JB  papafoad   jo  qoui 

ajBnbs  jad   spunod   (Xft'oi 

SuipuodsajjOD  ajnssaad  [Bixy 


OfNOfMOGOGOCO 


}0 

|O  qoui 

gjBnbs   J3d   spunod   QOg'oj 
SuipuodsajjoD  ajnssaid  [Bixy 


appwauBj.  j  ggggggsggsssgggg 


OOO'-iTH!-^i-li-ii3qs^S<>'N<»COCO-*J<-<t 

ooooooooooooooooo 


•pajapisuoo  jou 


8§8§§8j_8jjsli 

^  -H  o  o  cc  tr  Q  -cc  cc  o  o  co  g-g  cc  • «  o  ^ 


•qOUI  UB   JO  SIBIUID3Q  UI 

pox  2uipB3jqx  P  ssau^oi 


•M3JOC    10  -UIETQ    SB  SuOT  SB      I     ^^^         J^t         J?^         .^  ^<! 

'-5IMJ  jni  B  UT  -""",,    ,n  -n^T    I     COCDi-XOOOiGOOO 


— t  -^  i-i  (N  (N  CC 


JO  J3J3UIBIQ 


CC 

o  o  o_o  o  1-1 
tl 


^CO'^1~tlOOiXCO'*'^tl'*lOQO 
^  »0  O  O  O  O  <M  t-  rJJ  X •  CC  O  O 
i-  CC  »C  X  5  -t  «C  X  O  CC  (N  T- 


314 


MECHANICS. 


The  table  on  page  313  was  calculated   by   the  following 
formulas : 

When  friction  is  not  considered : 
Force  to  balance  load  =    Pjtch  m  m|hes  =    Pitch^inches 

When  both  one  pound  of  load  and  friction  are  considered, 


p  x  pitch  in  inches  -f-  middle  circum.  X  f  \    ^     /  middle  radius  \ 

^•middle  circum.  —  pitch  in  inches  X  f 


12 


CALCULATIONS  BY  TABLE  ON  PRECEDING  PAGE. 
EXAMPLE  1. 

A  jack  screw,  as  shown  in 
Fig.  19,  is  1*4"  diameter,  three 
threads  per  inch.  What  tan- 
gential force  is  required  to  act 
with  a  leverage  of  18  inches  in 
order  to  lift  5000  pounds  ?  Co- 
efficient of  friction  in  the  thread 
is  assumed  as  0.16.  Tangen- 
tial force  absorbed  in  friction 
by  the  collar  at  a  is  assumed 
to  be  equal  to  force  absorbed 
by  friction  in  the  thread  of 
the  screw,  and  may,  therefore, 
be  taken  from  the  thirteenth 
column  in  the  table. 
Solution : 

Tangential  force  per  pound  at  1  foot  radius  =  0.0133 
Tangential  force  absorbed  by  friction  in  collar  =  0.0089 
Total  force  per  pound  of  load  at  1  foot  radius  =  0.0222 
The  tangential  force  is  acting  with  18  inches  leverage  = 
1)4  feetj  and  the  load  is  5000  pounds;    therefore,  the  required 
force  will  be, 


F  — 


0.0222  X  5000 


—  74  pounds. 


EXAMPLE  2. 

A  load  of  16,000  pounds  rests  on  a  slide  and  is  moved  back 
and  forth  on  a  horizontal  plane  by  a  screw.  The  coefficient  of 
friction  between  slide  and  plane  is  0.1,  and  the  screw  should 
not  be  loaded  with  more  than  400  pounds  per  square  inch  of 
projected  area  or  thread.  Find  the  suitable  diameter  of  screw. 
If  a  pulley  of  20-inch  diameter  is  attached  to  the  end  of  the 
screw,  also  find  the  tangential  force  required  to  act  at  the  rim 
of  the  pulley  in  order  to  turn  the  screw. 


MECHANICS.  315 

Solution : 

The  coefficient  of  friction  for  the  slide  is  0.10,  therefore  the 
axial  pressure  on  the  screw  will  be  16,000  X  Vio  =  1600  pounds. 
The  allowable  force  on  a  1  j^-inch  screw  will  be  found  in  the 
table  to  be  1742  pounds ;  therefore,  select  a  screw  of  1^  inches 
diameter  and  a  length  of  nut  of  2^  inches.  Assuming  the 
friction  due  to  the  reaction  of  the  screw  against  its  collar  and 
bearing  to  be  equal  to  the  friction  in  the  thread,  and  using  the 
table,  we  have : 

Force  per  pound  at  one  foot  radius    =  0.0112 
Force  absorbed  by  friction  in  collar  =  0.0074 

Total  force  per  pound  of  load  at  one  foot  radius  =  0.0186 

The  leverage  of  a  20-inch  pulley  is  10  inches  =  ^As  foot, 
and  the  axial  force  is  1(500  pounds ;  therefore,  the  tangential 
force  required  at  the  rim  of  the  pulley  will  be  : 

F~    °-0186  X  160°    =36.7  pounds. 
*%» 

36.7  pounds  is  really  the  force  required  to  keep  the  body  in 
motion  after  it  is  started.  To  start  the  body  from  rest  requires 
somewhat  additional  force,  depending  on  the  time  used  in  over- 
coming its  inertia.  It  is  not  certain  that  the  friction  due  to  the 
reaction  of  the  screw  against  the  collar  is  equal  to  the  friction 
in  the  screw.  It  may  be  more  or  it  may  be  less ;  this  will,  to  a 
certain  extent,  depend  on  the  size  of  the  collar,  and  also  on  the 
finish  of  its  surfaces,  its  means  of  lubrication,  etc.  Therefore, 
instead  of 'assuming  this  resistance  to  be  equal  to  the  friction  of 
the  thread  as  found  in  column  10,  it  may  be  calculated  for 
each  individual  case  by  assuming  a  proper  coefficient  of  friction 
and  assuming  that  this  friction  acts  as  resistance  at  a  radius 
equal  to  the  middle  radius  of  the  collar.  If  a  screw  is  acting 
under  the  circumstances  illustrated  in  Fig.  18,  there  is  no  collar 
to  absorb  any  of  the  force  by  friction ;  but  whenever  the  screw 
acts  against  a  shoulder  this  friction  must  never  be  forgotten  in 
calculation.  Ball  bearings  may  be  used  to  very  good  advan- 
tage in  the  thrust  collar  on  a  screw.  If  a  screw  works  a  load 
continuously  up  and  down,  and  the  weight  of  the  load  always 
rests  on  the  screw,  it  is  necessary  to  be  very  careful  and  allow 
only  a  limited  load  on  the  screw  (only  a  fraction  of  what  is  given 
in  the  table),  because  the  pressure  of  the  load  always  acts  on 
the  same  side  of  the  thread,  and  this  is  very  disadvantageous 
for  lubrication,  as  it  does  not  give  the  oil  a  good  chance  to  get 
onto  the  surfaces  which  rub  against  each  other;  but  when  the 
screw  works  a  slide  with  an  alternate  push  and  pull,  the  wear 
comes  on  both  sides  of  the  thread,  which  gives  a  good  chance 
for  lubrication,  and  an  axial  pressure  of  400  pounds  per  square 
inch  of  projected  area  of  bearing  surface  in  the  thread  will  be 


3 1 6  MECHANICS. 

safe,  although,  under  certain  circumstances,  for  instance,  in  a 
mechanism  working  continuously,  such  a  load  may  be  too  much 
for  the  best  results  with  regard  to  wear. 

For  anything  working  like  a  jack-screw,  when  the  diameter 
of  the  screw  is  over  one  inch,  the  load  given  in  the  last  column 
is  perfectly  safe.  It  is  impossible  to  give  rules  which  will  suit 
all  cases ;  the  experience  and  judgment  of  the  designer  are  the 
best  guide  with  regard  to  the  selection  of  the  proper  load.  It 
may  seem  too  much  to  use  0.16  as  the  coefficient  of  friction  in 
the  thread  of  the  screw,  but  the  author  believes,  from  careful 
experiments  made  on  common  square-thread  screws,  as  used  in 
commercial  machinery, — not  made  for  experimental  purposes, 
but  for  every-day  use, — that  this  coefficient  of  friction  is  a  safe 
average.  It  is  well  to  remember  that  the  surfaces  of  the  thread 
on  screws  with  cast-iron  nuts  do  not  always  have  the  best  of 
finish,  and  the  nut  especially  is  liable  to  be  a  little  rough  when 
new  ;  therefore,  this  coefficient  of  friction  may  be  a  little  greater 
than  that  found  in  screws  in  machinery  when  well  lubricated 
and  with  surfaces  smoothed  down  and  glazed  over  from  wear. 

The  Parallelogram  of  Forces. 

A  line  may  be  drawn  to  such  FIG.  20. 

scale  that  its  length  represents  a 
given  force  acting  in  the  direction 
of  the  line.  Another  line  is  drawn 
to  the  same  scale,  from  the  same 
point  of  application,  and  its  length 
represents  another  force  acting  in 
the  same  direction  as  this  line.  If 
these  two  lines  are  connected  by 
two  auxiliary  lines,  a  parallelogram 
is  formed  and  the  diagonal  of  the  parallelogram  will  represent 
both  the  magnitude  and  the  direction  of  the  resulting  force. 

EXAMPLE. 

Let  the  lines  a  and  b  in  Fig.  20  represent  two  forces  acting 
in  the  direction  of  the  arrows.  Draw  the  lines  to  any  scale,  for 
instance,  TV  inch  to  a  pound ;  if  the  force  represented  by  a  is  64 
pounds,  the  line  a  will  be  64  X  TV  =  4"  long.  If  the  force  rep- 
resented by  b  is  50  pounds,  this  line  will  be  50  X  TV  =  3>£" 
long.  Completing  the  parallelogram  by  drawing  lines  c  and  dy 
the  diagonal,  x,  will  indicate  the  magnitude  and  direction  of  the 
resulting  force.  Suppose  these  two  forces  act  in  such  direc- 
tions that  when  the  parallelogram  is  completed  and  the  diagonal 
drawn,  it  is,  by  measurement,  found  to  be  4%."  long  =  |f; 
then  the  result  of  the  two  forces,  a  and  b,  is  a  force  of  76 
pounds.  In  many  cases,  the  result  of  force  and  stress  in  ma- 
chinery and  structures  may  very  conveniently  be  obtained  in  this 
way  with  much  less  labor  than  by  calculation,  and  with  accuracy 
consistent  with  good,  legitimate  practice, 


HORSE-  POWER.  317 

HORSE-POWER. 

The  term  horse-power,  as  applied  in  mechanical  calcula' 
tions,  is  33,000  foot-pounds  of  work  performed  per  minute,  or 
550  foot-pounds  of  work  per  second. 

To  Calculate  the  Horse-Power  of  a  Steam  Engine. 

RULE. 

Multiply  the  area  of  piston  in  square  inches  by  the  mean 
effective  steam  pressure,  and  this  by  the  piston  speed  in  feet 
per  minute,  and  divide  this  product  by  33,000.  The  quotient  is 
the  horse-power  of  the  engine. 

Formula  : 

Horse-power  =  0.7854  Z*  X   »  X  8  ,  X  » 


33000 

D  =  Diameter  of  piston  in  inches. 
p  =  Mean  effective  steam  pressure  in  pounds  per  square 

inch. 

s  =  Length  of  stroke  in  feet. 
n  =  Number  of  revolutions  per  minute. 

EXAMPLE. 

What  is  the  horse-power  of  a  steam  engine  of  the  following 
dimensions  ? 

Cylinder,  20  inches  diameter  ;  length  of  stroke,  3  feet  ; 
number  of  revolutions  per  minute,  75  ;  mean  effective  steam 
pressure  in  cylinder  during  the  stroke,  60  pounds  per  square 
inch. 

Horse-power  =  20*  X  °-7854  X  2  X  3  X  75  X  60 


Horse-power  = 


33000 
314.16  X  450  X  60 


33000 
Horse-power  —  257.04 


To  Calculate  the  Horse=Power  of  a  Compound  or  Triple 
Expansion  Engine. 

RULE. 

Calculate  the  mean  effective  pressure  of  the  steam  (accord- 
ing to  its  number  of  expansions  and  initial  pressure),  and  cal- 
culate the  horse-power  exactly  as  if  it  was  a  single  cylinder 
engine  of  the  same  size  as  the  size  of  the  last  cylinder. 

Another  way  is  to  take  indicator  diagrams  of  each  cylinder, 
and  calculate  the  power  of  each  cylinder  separately. 


318  HORSE-POWER. 

To  Judge  Approximately  the  Horse=Power  which  may  be 
Developed  by  Any  Common  Single  Cylinder  Engine. 

RULE. 

Square  the  diameter  of  the  piston  in  inches  and  divide 
by  2 ;  the  quotient  is  the  horse-power  which  the  engine  may 
develop. 

NOTE. — This  rule  gives  the  exact  horse-power,  if  the  prod- 
uct of  the  piston  speed  in  feet  and  the  average  pressure  per 
square  inch  in  the  cylinder  is  21,000. 

Horse=Power  of  Waterfalls. 

RULE. 

Multiply  the  quantity  of  water  in  cubic  feet  falling  in  a 
minute  by  62.5 ;  and  multiply  this  by  the  height  of  the  fall  in 
feet;  divide  this  product  by  33,000,  and  the  quotient  is  the 
horse-power  of  the  waterfall.  Or,  multiply  the  quantity  of 
water  in  cubic  meters  falling  in  a  minute,  by  1000,  and  multiply 
this  by  the  height  of  the  fall  in  meters ;  divide  the  product  by 
4500,  and  the  quotient  is  the  horse-power  of  the  waterfall. 

NOTE. — The  above  rules  give  the  gross  power  of  the  water- 
fall, but  the  useful  effect  of  the  fall  is  a  great  deal  less  and  will 
depend  on  the  construction  of  the  motor.  It  may  be  only  from 
40%  to  80%  of  the  natural  power  of  the  waterfall. 

Animal  Power. 

Under  favorable  circumstances,  a  horse  can  perform  22,000 
foot-pounds  of  work  per  minute.  For  instance,  a  horse  walking 
in  a  circle  turning  the  lever  in  a  so-called  horse-power  may 
exert  a  pull  of  100  pounds,  walking  at  a  speed  of  220  feet  per 
minute.  For  the  horse  to  work  to  advantage,  the  diameter  of 
the  circle  ought  to  be  at  least  25  feet. 

Hauling  a  Load. 

The  average  speed  when  horses  are  used  in  hauling  a  load 
one  way  and  returning  without  load  the  other  way,  allowing  for 
necessary  stoppages,  may  not  be  more  than  175  feet  per  minute, 
and,  in  estimating,  time  must  also  be  allowed  for  loading  and 
unloading.  Loads  may  vary  from  1000  to  2000  pounds,  accord- 
ing to  the  road.  Commonly  speaking,  the  force  required  to 
pull  a  loaded  wagon  on  a  good,  level  road  increases  in  propor- 
tion to  the  load  and  decreases  in  proportion  to  the  diameter  of 
the  wheels,  and  on  soft  roads  it  is  less  with  wide  tires  than  with 
narrow  ones.  The  idea  that  a  wagon  having  small  wheels 
would  be  easier  to  pull  up-hill  than  one  having  larger  wheels  is 
a  fallacy. 


HORSE-POWER.  319 

Power  of  Man. 

A  man  may  be  able  to  do  work  at  a  rate  of  4000  foot- 
pounds per  minute  ;  for  instance,  in  turning  a  crank  on  a  crane 
or  derrick,  a  force  of  15  pounds  may  be  exerted  on  a  crank,  18 
inches  long  and,  with  30  turns  per  minute,  the  work  would  be 
4228  foot-pounds  per  minute. 

NOTE. — In  derricks,  pulley  blocks,  jack-screws,  etc.,  a  large 
part  of  the  expended  power  is  consumed  in  overcoming  friction. 

Power  Required  to  Drive    Various   Kinds   of  Machinery. 

In  the  nature  of  the  thing  it  is  impossible  from  experiments 
on  one  machine  to  tell  exactly  what  power  it  takes  to  run  an- 
other similar  machine,  as  there  are  so  many  different  factors 
entering  into  the  problem  ;  for  instance,  the  speed  and  feed  on 
the  machine,  the  hardness  of  the  stock  it  works  on,  the  quality 
of  the  tools  used,  the  kind  of  lubrication,  etc.  Therefore,  such 
assertions  are  only  approximations  at  the  best. 

16-inch  engine  lathe,  back  geared,  %  horse-power. 

26-inch  engine  lathe,  back  geared,  1  %  horse-power. 

Planer,  22"  x  22"  x  6  feet,  l/2  horse-power. 

Planer,  32"  x  32"  x  10  feet,  %  horse-power. 

Shaping  machine,  10-inch  stroke,  #  horse-power. 

20-inch  drill  press,  ]/2  horse-power. 

26-inch  drill  press,  back  gear,  boring 

a  3-inch  hole,  using  boring  bar,  1  horse-power. 

Plain  milling  machines  (Lincoln 

•pattern,  No.  2),  1%.  horse-power. 

Small  Universal  milling  machines,  %  horse-power. 

Circular  saws  (for  wood),  24"  di- 
ameter (light  work),  3>£  horse-power. 

Circular  saws  (for  wood),  36"  di- 
ameter (light  work),  6  horse-power. 

Fan  blower  for  cupola,  melting  four 

tons  of  iron  per  hour,  10  horse-power. 

Fan  blower  for  five  blacksmith  fires,  1      horse-power. 

Drop  hammer,  800  pounds,  8      horse-power. 

In  machine  shops  and  similar  places,  from  40%  to  70%  of 
the  total  power  required  is  consumed  in  running  the  line  shaft- 
ing and  counter-shafts.  An  average  of  from  55%  to  60%  is 
probably  the  most  common  ratio. 

In  exceptionally  well-arranged  establishments,  under  favor- 
able conditions,  in  light  manufacturing  it  may  be  possible  that 
only  30%  of  the  power  is  consumed  in  driving  line  and  counter 
shafting,  and  that  70%  is  used  for  actual  work. 


320 


SPEED   OF    MACHINERY. 


SPEED   OF   MACHINERY. 

The  peripheral  velocity  of  circular  saws  ought  not  to  exceed 
10,000  feet  per  minute.  Table  No.  37  gives  the  number  of  revo- 
lutions per  minute  for  circular  saws  of  different  diameters. 

TABLE  No.  37- 


Diameter  of 
saw  in  inches. 

8 

10 

12 

14 

16 

20 

24 

28 

32 

Number  of 
revolutions 
per  minute. 

4500 

3600 

3000 

2585 

2222 

1800 

1500 

1285 

1125 

Band  Saws. 

Small  band  saws,  such  as  are  usually  used  in  carpenter 
shops,  have  a  velocity  of  3600  feet  per  minute.  The  reason 
why  band  saws  are  run  so  much  slower  than  circular  saws  is 
that  if  the  band  saw  is  given  too  much  speed  the  blade  will  be 
pulled  to  pieces  in  starting  and  stopping. 

Drilling  flachines  for  Iron. 

For  drilling  steel,  the  surface  speed  of  a  drill  should  not 
exceed  15  feet  per  minute;  cast-iron,  22  feet;  brass,  27  feet; 
malleable  iron,  25  to  30  feet  per  minute.  The  feed  will  vary 
according  to  the  hardness  of  the  stock.  In  cast-iron  a  %"  drill 
will  drill  a  hole  1"  deep  in  125  revolutions.  A  l/2"  drill  will 
drill  a  hole  I"  deep  in  120  revolutions.  A  1"  drill  will  drill  a 
hole  1"  deep  in  100  revolutions. 

Lathes. 

Cast-iron  may  be  turned  at  a  speed  of  32  feet  per  minute 
when  Muchet  steel  is  used  for  tools.  Thus,  lathes  are  usually 
calculated  to  have  a  velocity  of  about  30  to  32  feet  on  the  slovy- 
est  speed,  supposing  that  as  large  a  diameter  as  the  lathe  will 
swing  is  turned. 

For  wood-turning  the  surface  speed  may  be  from  3000  to 
0000  feet  per  minute ;  but  when  the  article  to  be  turned  is  out  of 
balance  the  speed  must  be  considerably  slower. 

Planers. 

Cast-iron  is  planed  at  a  speed  of  25  to  27  feet  per  minute ; 
wrought  iron,  21  feet;  steel.  16  feet  per  minute.  A  planer  ought 
to  return  at  least  three  times  as  fast  as  it  goes  forward. 

Hilling  Machines. 

Rotating  cutters  working  on  Bessemer  steel  or  other  mate- 
rials of  about  equal  hardness  usually  have  a  surface  speed  of 


SPEED    OF    MACHINERY.  321 

about  40  feet  per  minute.     Oil  is  used  for  lubrication.     Cast- 
iron  is  milled  without  oil. 

Grindstones. 

When  grindstones  are  used  to  grind  steel  and  iron  in  manu- 
facturing, they  work  at  a  surface  speed  of  2000  to  2500  feet  per 
minute,  but  grindstones  for  common  shop  use,  to  grind  tools, 
chisels,  etc.,  run  at  much  slower  speed. 

Emery  Wheels  and  Emery  Straps. 

Emery  wheels  and  straps  do  good  work  at  a  speed  of  5000 
to  6000  feet  per  minute,  but  all  such  high-speed  machinery, 
especially  grindstones  and  emery  wheels,  must  be  used  very 
carefully  and  special  attention  paid  to  the  strength,  so  that  they 
will  not  break  under  the  stress  of  centrifugal  force. 

Calculating  Size  of  Pulleys. 

TO   FIND   SIZE  OF   PULLEY   ON   MAIN   SHAFT. 

Multiply  the  diameter  of  pulley  on  counter-shaft  by  its 
number  of  revolutions  per  minute,  and  divide  this  product  by 
the  number  of  revolutions  of  the  main  shaft,  and  the  quotient  is 
the  diameter  of  the  pulley  on  the  main  shaft. 

EXAMPLE. 

A  main  shaft  makes  150  revolutions  per  minute ;  the  counter- 
shaft has  a  pulley  9  inches  in  diameter  and  is  to  make  400  revolu- 
tions per  minute.  What  size  of  pulley  is  required  on  the  main 
shaft? 

Solution : 

Diameter  of  pulley  —  40°  *  9  =  24  inches. 
150 

TO   FIND   SIZE   OF   PULLEY   ON   COUNTER-SHAFT. 

RULE. 

Multiply  the  diameter  of  pulley  on  the  main  shaft  by  its 
number  of  revolutions  per  minute,  and  divide  this  product  by 
the  number  of  revolutions  of  the  counter-shaft;  the  quotient  is 
the  diameter  of  the  pulley  on  the  counter-shaft. 

EXAMPLE. 

The  pulley  on  a  main  shaft  is  30  inches  in  diameter  and  it 
makes  150  revolutions  per  minute  ;  the  counter-shaft  is  to  make 
450  revolutions  per  minute.  What  size  of  pulley  is  required  ? 

Solution : 

Diameter  of  pulley  =   36  X  15°  =  12  inches. 


322  SPEED  OF  MACHINERY. 

TO  FIND  THE  NUMBER  OF  REVOLUTIONS  OF  THE  COUNTER 

SHAFT. 

RULE. 

Multiply  the  diameter  of  pulley  on  the  main  shaft  by  its 
number  of  revolutions  per  minute  and  divide  this  product  by 
the  diameter  of  pulley  on  the  counter-shaft,  and  the  quotient  is 
the  number  of  revolutions  of  the  counter-shaft  per  minute. 

EXAMPLE. 

The  pulley  on  a  main  shaft  is  24  inches  in  diameter  and 
makes  150  revolutions  per  minute,  and  the  pulley  on  the  counter- 
shaft is  15  inches  in  diameter.  How  many  revolutions  per 
minute  will  the  counter-shaft  make? 

Number  of  revolutions  =  =  240  revolutions  per  minute. 

15 

To  Calculate  the  Speed  of  Gearing. 

In  calculating  the  speed  of  gearing,  use  the  same  rules  as 
for  belting,  but  take  the  number  of  teeth  instead  of  the 
diameter. 

EXAMPLE. 

The  back  gearing  on  a  lathe  consists  of  a  gear  and  pinion 
of  8  pitch,  90  teeth  and  32  teeth,  and  the  other  gear  and  pinion 
are  10  pitch,  120  teeth  and  40  teeth.  How  many  revolutions  will 
the  cone  pulley  make  while  the  spindle  makes  one  revolution  ? 

Solution : 

Cone  pulley  makes  =   9^  X  12°    =  9  revolutions. 

o^£  /\  40 

Efficiency  of  Machinery. 

Divide  the  energy  given  out  by  a  machine  by  the  energy 
put  into  the  same  machine  ;  multiply  the  quotient  by  100,  and 
the  result  is  the  per  cent,  of  efficiency  of  the  machine. 


A  dynamo  requires  15  horse-power,  but  the  electrical  power 
given  out  is  only  12  horse-power.     What  is  the  efficiency? 
Solution : 

Efficiency  =  -If-  X  100  = 
15 

A  steam  engine  is  to  develop  60  horse-power  net.    What 
will  be  the  gross  horse-power  if  the  efficiency  is  75%  ? 
Solution : 

Gross  power  =  60  *  10°  =  80  horse-power. 
<5 


CRANE    HOOKS. 

CRANE  HOOKS. 


3*3 


FIG.  3. 


Crane  hooks,  as  shown  in  Figs.  1,  2  and  3,  may  be  designed 
by  the  following  formulas  : 


P  =  Load  in  tons. 

D  =  Diameter  of  iron  in  inches. 


d—\y2D 


S  —  Standard  screw  of  diameter  r 


*  —  Vl6  D 


When  a  rectangular  iron  plate  is  substituted 
for  a  washer,  the  bearing  surface  of  the  plate 
against  the  wood  should  at  least  be  equal  to 
the  area  of  the  washer,  calculated  by  the  above 
formula. 

Chain  Links. 

(See  Figure  4.) 

D  =  Diameter  of  iron. 
L  =  ±y2  to5Z>. 


(For  strength   of  chains,  see  page  222). 


FIG.  4. 


3*4 


CRANES  AND   DERRICKS. 


CRANES. 


Cranes  and  derricks  are 
machines  used  for  raising  and 
lowering  heavy  weights.  In 
its  simplest  form,  a  crane  con- 
sists of  three  principal  mem- 
bers :  The  upright  post,  the 
horizontal  jib  and  the  diagonal 
brace.  (See  Fig.  5).  The 
weight  P  will  produce  tensile 
stress  in  the  jib,  compressive 
stress  in  the  brace,  and  both 
compressive  and  transverse 
stress  in  the  post. 

Tension  in  jib  =  P  X  x 


FIG.  5. 


Compression  in  brace  =  — — - 

y 


.  PKk 


Stress  in  the  upper  bearing  = 

e 

When  the  post  is  held  at  both  ends,  as  in  Fig.  5,  it  may, 
with  regard  to  transverse  strength,  be  considered  as  a  beam  of 
length  t,  fastened  at  one  end  and  loaded  at  the  other  with  a  load 


equal  to  the  force 


h  X  P 


The  compression  on  the  post  caused  by  the  load  is  equal 
to  />. 

The  downward  pressure  on  the  lower  bearing  is  equal  to 
the  sum  of  the  weight  of  the  crane  and  the  load  which  it 
supports. 


Proportions  for  a  Two-Ton  Derrick 

(Of  the  construction  shown  in  Fig.  6) . 

Pulley  blocks  should  be  double-sheave  (only  single  are 
shown  in  the  cut).  Circumference  of  manila  rope,  3#  inches. 
Mast,  8X8  inches,  26  feet  long.  Boom,  7X7  inches,  20  feet  long. 


CRANES  AND   DERRICKS. 


325 


FIG.  6. 


Large  gear,  72  teeth,  1- 
inch  circular  pitch,  2-inch 
face.  Small  pinion,  12 
teeth,  1-inch  circular  pitch, 
2-inch  face.  Crank  shaft. 
\l/2  inches  in  diameter. 
Bearings,  2^  inches  long. 
Crank,  18  inches  long, 
Drum,  7  inches  in  diame- 
ter, 24  inches  long.  Drum- 
shaft,  2X  inches  in  di- 
ameter. The  drum  and 
large  gear  are  fitted  and 
keyed  to  the  drum  shaft 
and  also  bolted  together, 
thereby  relieving  this  shaft 
from  twisting  stress. 

The  radius  of  the 
drum  added  to  the  radius 
of  the  rope  makes  four 
inches,  and  the  force  is 
multiplied  five  times  by 
the  double-sheave  pulley 
block;  therefore,  when  Gear 
the  friction  in  thecrank 
mechanism  is  not  consid- 
ered, the  force  required 
on  the  crank  in  order  to 
lift  4400  pounds  will  be : 

F  =  -_LX_12  X  440°     —  33  pounds,  very  nearly. 
18  X  72  X  5 

Thus,  when  two  men  are  working  the  derrick  (one  at  each 
crank),  each  man  has  to  exert  a  force  of  \Ql/2  pounds, 
but,  including  friction,  each  man  probably  exerts  a  force  of  20 
to  25  pounds,  when  the  derrick  is  loaded  to  its  full  capacity. 

For  very  rapid  work  it  is  necessary  to  have  four  men  (two 
on  each  winch-handle)  to  work  the  derrick,  if  it  is  kept  loaded 
to  its  maximum  capacity,  but  for  ordinary  stone  work  such  a 
derrick  is  usually  worked  by  two  men.  Stones  as  heavy 
as  two  tons  are  seldom  handled,  except  where  larger  derricks 
and  steam  power  are  used. 

When  the  derrick  is  to  be  worked  constantly,  the  limit  of 
the  average  stress  on  the  crank  handle  to  be  allowed  for  each 
man  is  15  pounds.  When  working  an  18-inch  crank,  48  turns 
per  minute,  this  corresponds  to  a  force  of  15  pounds  acting 
through  a  space  of  a  little  over  220  feet  —  3300  foot-pounds  of 
work  per  minute  =  jV  horse-power. 

When  the  crank  swings  in  a  shorter  radius  a  few  more 
turns  per  minute  may  be  expected,  but  experience  indicates  that 
an  18"  radius  is  the  most  practical  proportion. 


326  BELTS. 

BELTS. 

Oak-tanned  leather  is  considered  the  best  for  belting.  The 
so-called  "short  lap"  is  cut  lengthwise  from  the  middle  of  the 
back  of  the  hide,  where  it  has  the  most  firmness  and  strength. 
Single  belting  more  than  three  inches  in  width  is  about  Ty 
thick,  and  weighs  15  to  16  ounces  per  square  foot j  when  less 
than  three  inches  in  width  it  is  usually  -f.,"  thick  and  weighs 
about  13  ounces  to  the  square  foot. 

Light  double  belts,  as  used  for  dynamos  and  other  ma- 
chinery having  pulleys  of  comparatively  small  diameter,  are 
about  ¥y  thick  and  weigh  about  21  ounces  per  square  foot. 
Double  "belting,  as  used  for  main  belts,  is  a  little  heavier  and 
weighs  from  25  to  28  ounces  per  square  foot.  Belts  as  heavy  as 
30  ounces  per  square  foot  are  frequently  used,  and  are  usually 
termed  "heavy  double."  Large  engine  belts  are  sometimes 
made  with  three  thicknesses  of  leather. 

Belts  should  be  soft,  pliable  and  of  even  thickness.  When 
a  belt  is  of  uneven  thickness  and  has  very  long  joints,  so  that 
it  looks  as  if  it  was  partly  single  and  partly  double,  it  is  very 
doubtful  if  it  will  do  good  service,  for  this  is  a  sure  sign  that  the 
thin  and  flimsy  parts  of  the  hide  have  been  taken  into  the  stock 
in  making  the  belt. 

The  ultimate  tensile  strength  of  leather  belting  is  from  2600 
to  4800  pounds  per  square  inch  of  section.  Thus,  a  leather  belt 
Ty  thick  will  break  at  a  stress  of  500  to  900  pounds  per  inch  of 
width. 

The  lacing  of  belts  will  reduce  their  strength  from  50  to  60 
per  cent.;  therefore,  when  practicable,  belts  ought  to  be  made 
endless  by  cementing  instead  of  lacing. 

A  belt  will  transmit  more  power,  wear  better  and  last 
longer,  if  it  is  run  with  the  grain  side  next  to  the  pulley. 

Belts  should  never  be  tighter  than  is  necessary  in  order  to 
transmit  the  power  without  undue  slipping ;  too  tight  belts  cause 
hot  bearings,  excessive  wear  and  tear,  and  loss  of  power  in  over- 
coming friction ;  but,  on  the  other  hand,  it  is  necessary  to  have  a 
belt  tight  enough  to  prevent  it  from  slipping  on  the  pulley,  be- 
cause if  a  belt  slips  there  is  not  only  a  direct  loss  in  velocity,  but 
the,  belt  will  wear  out  in  a  short  time ;  it  is,  therefore,  very  im- 
portant to  use  belts  of  such  proportions  that  the  power  shall  be 
transmitted  with  ease. 

Belts  always  run  toward  the  side  of  the  pulley  which  is 
largest  in  diameter  (therefore  pulleys  are  crowned,  in  order  to 
keep  the  belt  running  straight). 

A  belt  will  always  run  toward  the  side  where  the  centers  of 
the  shafts  are  nearest  together. 

Open  belts  will  cause  two  shafts  to  run  in  the  same  direction, 


BELTS. 


327 


A  crossed  belt  will  cause  the  shafts  to  run  in  opposite  direc- 
tions. If  the  distance  between  the  shafts  is' short,  crossed  belts 
will  not  work  well.  A  short  belt  will  wear  out  faster  than  a 
long  one. 

Very  long  and  heavy  belts  should  be  supported  by  idlers  as 
well  under  the  slack  as  under  the  working  side ;  if  not,  the 
weight  of  a  long  belt  will  cause  too  much  stress  on  itself  and 
also  cause  too  much  pressure  on  the  bearings,  as  well  on  the 
driver  as  on  the  driven  shaft.  Belts  should  never,  when  it  can 
be  avoided,  be  run  vertically,  as  the  weight  of  the  belt  always 
tends  to  keep  it  away  from  the  lower  pulley,  thereby  reducing 
its  transmitting  capacity;  the  longer  the  belt  the  worse  this  is. 
Belts  are  most  effective  when  they  are  run  in  a  horizontal  direc- 
tion and,  whenever  possible,  the  lower  part  of  the  belt  should  be 
the  working  part,  as  the  slackness  in  the  upper  part,  by  its 
weight,  will  cause  the  belt  to  lay  around  the  pulley  for  a  longer 
distance,  and  this  will,  in  a  measure,  increase  its  transmitting 
capacity;  but  if  the  upper  part  is  the  working  part,  the  slackness 
in  the  lower  part  tends  to  keep  the  belt  away  from  the  pulleys, 
and  thereby  reduces  its  transmitting  capacity. 


Lacing  Belts. 

Figure  1  shows  a  good  way  of  lacing  belts ;  a  is  the  side  run- 
ning next  to  the  pulley  and  b  is  the  outside.  Holes  should  be 
punched  and  not  made  by  an  awl,  as  punched  holes  are  less  lia- 
ble to  tear.  The  lacing  is  commenced  by  putting  each  end  of 
the  lace  through  holes  1  and  2  from  the  side  next  to  pulley,  and 
then  continuing  toward  the  edges,  both  sides  simultaneously, 

FIG.  1. 


a 


w\ 

JL 


making  a  double  stitch  at  the  edges  and  sewing  back  again  un- 
til holes  1  and  2  are  reached ;  and,  lastly,  by  drawing  each  end 
of  the  lace  through  .r  and  y.  Each  stitch  w'ill  be  double,  except- 


328  BELTS. 

ing-  the  middle  one.  The  holes  x  and  y,  where  the  ends  of  the 
lacing  are  finally  drawn  through  for  fastening,  are  made  by  the 
belt  awl  and  should  always  be  made  small,  and  the  lacing,  if  laid 
out  rightly,  always  enters  these  holes  from  the  inside  of  the  belt ; 
after  it  is  pulled  through,  a  small  cut  is  made  in  the  lacing  on 
the  outside,  which  will  prevent  it  from  drawing  back  again,  then 
the  ends  are  cut  off  about  l/2"  long,  as  shown  in  the  figure  at.r 
and  j>.  It  is  a  bad  practice  to  leave  the  lace-ends  on  the  inside  of 
belts,  because  they  will  then  soon  wear  off,  allowing  the  joint  to 
rip. 

A  1-inch  belt  ought  to  have  three  lace-holes  in  each  end. 
Length  of  lacing,  12  inches. 

A  2-inch  belt  ought  to  have  three  lace-holes  in  each  end. 
Length  of  lacing,  18  inches. 

A  3-inch  belt  ought  to  have  five  lace-holes  in  each  end. 
Length  of  lacing,  24  inches. 

A  4-inch  belt  ought  to  have  five  lace-holes  in  each  end. 
Length  of  lacing,  32  inches. 

A  5-inch  belt  ought  to  have  seven  lace-holes  in  each  end. 
Length  of  lacing,  40  inches. 

A  6-inch  belt  ought  to  have  seven  lace-holes  in  each  end. 
Length  of  lacing,  48  inches. 

An  8-inch  belt  ought  to  have  nine  lace-holes  in  each  end. 
Length  of  lacing,  60  inches. 

A  10-inch  belt  ought  to  have  eleven  lace-holes  in  each  end. 
Length  of  lacing,  72  inches. 

A  12-inch  belt  ought  to  have  thirteen  lace-holes  in  each  end. 
Length  of  lacing,  84  inches. 

Always  have  the  row  having  the  most  holes  nearest  the  end 
of  the  belt. 

Cementing  Belts. 

When  belts  are  cemented  together,  a  3-inch  belt  is  lapped 
four  inches  and  a  4-inch  belt  4^  inches.  In  larger  belts  the  lap 
is  usually  made  equal  to  the  width  of  the  belt,  but  it  may  be 
made  even  shorter  when  the  width  of  the  belt  is  over  12  inches. 
The  two  ends  are  jointed  together,  so  that  the  thickness  is  even 
with  the  rest  of  the  belt. 

The  American  Machinist,  in  answer  to  Question  No.  430, 
Dec.  5, 1895,  says :  "  For  leather  belts  take  of  common  glue  and 
American  isinglass  equal  parts;  place  them  in  a  glue  pot  and 
add  water  sufficient  to  just  cover  the  whole.  Let  it  soak  10 
hours,  then, bring  the  whole  to  a  boiling  heat,  and  add  pure  tan- 
nin until  the  whole  appears  like  the  white  of  an  egg.  Apply 
warm.  Buff  the  grain  of  the  leather  where  it  is  to  be  cemented; 
rub  the  joint  surfaces  solidly  together,  let  it  dry  for  a  few  hours, 
and  the  belt  will  be  ready  for  use.  For  rubber  belts  take  16 
parts  gutta  percha,  4  parts  India  rubber,  2  parts  common  caulk- 
er's pitch,  1  part  linseed  oil ;  melt  together  and  use  hot.  This 
cement  can  also  be  used  for  leather," 


BELTS.  329 

Length  of  Belts. 

Small  belts,  such  as  4  inches  wide  or  less,  will  work  well 
when  the  distance  between  the  shafts  is  from  12  to  15  feet,  larger 
belts  when  from  20  to  25  feet,  and  for  large  main  belts  25  to  30 
feet  distance  is  satisfactory. 

Horse=Power  Transmitted  by  Belting. 

A  single  belt  weighing  about  15  ounces  per  square  foot  is 
capable  of  transmitting  one  horse-power  per  inch  of  width, 
when  running  at  a  speed  .of  800  feet  per  minute  over  pulleys  of 
proper  size,  both  of  equal  diameter.  As  one  horse-power  is 
33,000  foot-pounds  of  work  per  minute,  this  will  make  the  tension 

38000 
due  to  the  power  the  belt  is  transmitting  =      ^     '=  41  y^  Ibs. 

per  inch  of  width,  but  the  total  tension  in  the  belt  is,  of  course, 
considerably  more  per  inch  of  width,  because  the  belt  must  be 
tight  enough  to  prevent  its  slipping  on  the  pulley.  For  belts 
lighter  than  15  ounces  per  square  foot  it  is  better  to  allow  1000 
running  feet  per  horse-power  per  inch  of  width  of  belt.  For  light 
double  belts  weighing  21  ounces  per  square  foot,  600  running 
feet  per  horse-power  per  inch  of  width  may  be  allowed.  For 
double  belts  weighing  25  ounces  per  square  foot,  500  running 
feet  per  horse-power  per  inch  of  width  may  be  allowed.  Hence 
the  following  formulas  : 

For  light  single  belts  weighing  less  than  15  ounces  per 
square  foot, 

TT  _      v  X  b  JJX  1000 

~~Iobo~  ~^~ 

For  single  belts  weighing  15  to  16  ounces  per  square  foot, 
H  -v  X  b  HX  800 

800""  v 

For  light  double  belts  weighing  about  21  ounces  per  square 
foot, 

7,X*  //X600 

<•><)<)  v 

For  double  belts  weighing  about  25  ounces  per  square  foot, 

ff  -      ""  X  b  t=    //X500 

500  v 

H  =  Horse-power. 
b  —  Width  of  belt  in  inches. 

•v  —  Velocity  of  belt  in  feet  per  minute,  which  will  be  di- 
ameter of  pulley  in  inches  multiplied  by  3.1416  and  by  the  num- 
ber of  revolutions  per  minute,  and  the  product  divided  by  12, 


33°  BELTS. 

EXAMPLE  1. 

A  double  belt  10  inches  wide,  weighing  25  ounces  per  square 
foot,  runs  over  50-inch  pulleys,  making  240  revolutions  per  min- 
ute. How  many  horse-power  will  it  properly  transmit  ? 

Solution  : 

'.     .        ,  .    ,          50  X  3.1416  X  240 
Velocity  of  belt  =  -  12  -   =  3141.6  ft.  per  minute. 

3141.6  X  10 
~        —  500  --  =  ^2'^  norse'P°wer- 

EXAMPLE  2. 

One  hundred  horse-power  is  to  be  transmitted  by  a  double 
belt  weighing  25  ounces  per  square  foot.  The  pulleys  are  66 
inches  in  diameter  and  make  150  revolutions  per  minute.  What 
is  the  necessary  width  of  belt  ? 

Solution  : 

Pulleys  of  66  inches  diameter,  running  150  revolutions  per 
minute,  will  give  a  belt  speed  of  15°  X  3'1416  X  66  _  3591.8; 

say,  2592  feet  per  minute. 

100  X  500 
—  —  2592  -  —    19-3  inches;    thus,  a  double  belt  20 

inches  wide  will  do  the  work. 

EXAMPLE  3. 

A  light  single  belt  4  inches  wide,  weighing  13  ounces  per 
square  foot,  runs  over  pulleys  of  36  inches  diameter,  making  100 
revolutions  per  minute.  How  many  horse-power  may  be  trans- 
mitted? 

Solution  : 

36  X  3.1416  X  100 
Velocity  of  belt  =  -        —  ^  —         -  =  942.48  ft.  per  minute. 

The  belt  is  a  light  single  belt  and  its  transmitting  capacity 

4  X  942.48 
will  be,  H  =  --        --    =  3.76992,  about  3^  horse-power. 


To  Calculate  Size  of   Belt  for   Given  Horse-Power  when 

Diameter  of  Pulley  and  Number  of  Revolutions 

of  Shaft  Are  Known. 

The  following  formulas  may  be  used  for  calculating  belt 
transmission,  and  will  give  results  approximately  consistent 
with  previously  given  rules,  but  they  are  more  convenient  for 
use,  as  the  velocity  of  the  belt  does  not  need  to  be  first  calculat- 
ed, but  the  velocity  of  the  belt  must  not  exceed  the  practical 
limit. 


BELTS.  331 

This  formula  will  do  for  either  single  or  double  leather  belts 
with  cemented  joints  (no  lacing),  of  any  weight  from  12  to  30 
ounces  per  square  foot  and  of  any  width  from  one  to  thirty 
inches,  when  the  pulleys  are  of  suitable  size  to  correspond  with 
the  thickness  of  the  belt,  and  the  diameter  of  both  pulleys  is 
equal  or  nearly  so : 

_     d  X  n  X  b  X  w  _       HX  50000 

~liOOOO~~  d  ~       nXbXw 

H  X  50000  _      H  X  50000  H  X  50000 


H  =  Horse-power  transmitted  by  the  belt. 

d  —  Diameter  of  pulley  in  inches. 

n  =  Number  of  revolutions  per  minute. 

b  =  Width  of  belt  in  inches. 
iv  =  Weight  of  belt  in  ounces  per  square  foot. 
50,000  is  constant. 

EXAMPLE. 

Calculate  Example  2  by  the  above  formula. 

Solution: 

100  X  50000 

^  ~  06  X  150~X~25  =  20'^  mcnes'  wnicn»  f°r  aU  practical 
purposes,  is  the  same  as  the  result  when  calculated  by  the  other 
rule. 

Wide  and  thin  belts  are  unsatisfactory.  It  is  far  better 
when  transmitting  power  to  use  double  and  narrow  rather 
than  single  and  wide  belts.  It  is  a  very  bad  practice  to  run 
at  too  slow  belt  speed,  and  also  to  use  pulleys  of  too  small  diam- 
eter. The  smallest  pulley  for  a  light  double  belt  should  never 
be  less  than  12"  in  diameter,  for  a  heavy  double  belt  never  less 
than  20"  in  diameter,  and  for  a  triple  belt  the  pulley  should  not 
be  less  than  30"  in  diameter. 

To  Calculate  Width  of  Belt  when  Pulleys  are  of  Unequal 
Diameter. 

When  the  pulleys  are  of  different  diameters  the  belt  will  lay 
around  the  smallest  pulley  less  than  ISO  degrees,  and  the  trans- 
mitting capacity  of  the  belt  is  correspondingly  reduced.  The 
pressure  on  the  pulley  due  to  the  tension  of  the  belt  will  vary  as 
the  sine  of  half  the  angle  of  contact,  and  the  adhesion  of  the  belt 
to  the  pulley  will  vary  as  the  pressure;  consequently,  also,  the 
transmitting  capacity  of  the  belt  will  vary  as  the  sine  of  half  of 
the  angle  of  contact,  but  it  is  usually  advisable  in  practice  to 
allow  a  little  more  on  the  width  of  the  belt  than  is  called  for 
by  this  rule.  A  practical  rule  is  : 

First  calculate  the  width  of  the  belt  by  the  above  rules 
and  formulas,  as  though  both  pulleys  had  the  same  diameter, 


332  BELTS. 

then  multiply  the  result  by  the  following  constants,  according  to 
the  arc  of  contact  between  the  belt  and  the  small  pulley. 

When  the  arc  of  contact  between  the  belt  and  the  small 
pulley  is  90°  multiply  by  1.60. 

100°        "          "     1.45  140°  multiply  by  1.15 

110°        "          "    1.35  150°        "          "    1.10 

120°        "          "     1.25  160°         "          "     1.06 

130°  "     1.20  170°        "          "    1.04 

EXAMPLE. 

The  pulley  on  a  dynamo  is  15"  in  diameter,  and  it  makes 
1200  revolutions  per  minute.  The  driving  pulley  is  so  large  that 
the  belt  only  lays  around  the  dynamo  pulley  for  a  distance  of 
150  degrees.  What  is  the  necessary  width  of  a  light  double  belt, 
weighing  21  ounces  per  square  foot,  when  it  takes  40  horse-power 
to  run  the  dynamo  ? 

Solution : 

If  the  arc  of  contact  had  been  180  degrees  the  belt  would 

40  X  50000 

be    b    =   1200  X  15~X~2T  :  :  inches  wide,  but  as  the  arc 

of  contact  is  not  180  degrees,  but  only  150  degrees,  this  width 
is  multiplied  by  the  constant  1.10,  as  given  in  the  preceding 
table.  Thus,  the  width  of  the  belt  will  be  5.3  X  1.1  =  5.83 
inches  or,  practically,  a  belt  six  inches  wide  is  required. 

When  belts  are  running  in  a  horizontal  direction, 
and  the  driven  pulley  and  the  driver  are  of  equal  diameter 
and  finish,  the  belt  will  always,  when  overloaded,  commence  to 
slip  on  the  driver,  and  when  pulleys  are  of  unequal  size  it  is 
always  more  favorable  for  the  belt  when  the  driving  pulley  is 
the  larger  than  when  vice  versa. 

To  Find  the  Arc  of  Contact  of  Belts. 

Make  a  scale  drawing  of  the  pulleys  and  the  belt,  and 
measure  the  arc  of  contact  from  the  drawing  by  means  of  a 
protractor,  or  the  arc  of  contact  in  degrees  on  the  small 
pulley  for  an  open  belt  may  be  calculated  by  the  formula: 

Cosine  of  half  the  angle  =  R~r 

R  =  Radius  of  large  pulley  in  inches. 
r  =  Radius  of  small  pulley  in  inches. 
/  =  Distance  in  inches  between  centers  of  the  shafts. 

EXAMPLE. 

The  distance  between  centers  of  two  shafts  is  16  feet ;  the 
large  pulley  is  60  inches  and  the  small  pulley  is  20  inches 
in  diameter.  What  is  the  arc  of  contact  of  the  belt? 


BELTS.  333 

Solution : 

16  feet  =  192  inches. 

60  inches  diameter  =  30  inches  radius. 

20  inches  diameter  =  10  inches  radius. 

Cos.  of  half  the  angle  =    (30  —  10)  —  0.104 

192 

In  tables  of  natural  cosine  (page  158),  the  corresponding 
angle  is  found  to  be  84  degrees,  very  nearly ;  thus,  the  angle 
for  arc  of  contact  will  be  2  X  84  =  168  degrees  on  the  small 
pulley.  On  the  large  pulley  the  arc  of  contact  will  be  360  — 
168  =  192  degrees. 

For  a  crossed  belt  the  arc  of  contact  is  always  the  same  on 
both  pulleys,  and  it  may  be  calculated  by  the  formula : 

Cos.  of  half  the  angle  =  —  — j — 

A*  =  Radius  of  large  pulley. 
r  —  Radius  of  small  pulley. 
/  =  Distance  between  centers. 
EXAMPLE. 

What  will  be  the  arc  of  contact  for  the  belt  on  the  pulleys 
in  the  previous  examples  if  belt  is  run  crossed  instead  of  open  ? 

Solution : 

Cosine  of  half  the  angle  =  —  30  +  10    =    —  0.208  ;     the 

192 

corresponding  angle  will  be  180  —  77  =  103  degrees,  and  the 
arc  of  contact  will  be  103  X  2  =  206  degrees. 

Pressure  on  the  Bearings  Caused  by  the  Belt. 

Approximately,  the  pressure  on  the  bearings  caused  by  the 
belt  may  be  considered  to  be  three  times  the  force  which  the 
belt  is  transmitting.  Therefore,  the  pressure  may  be  calculated 
by  the  formula : 

p  =  3  X  33000  X  H 
v 

P  =  Pressure  on  the  bearings  due  to  pull  of  belt. 

H  •=•  Number  of  horse-power  transmitted  by  the  belt. 

v  =  Velocity  of  belt  in  feet  per  minute. 

EXAMPLE  1. 

A  belt  is  transmitting  60  horse-power  and  its  velocity  is  900 
feet  per  minute.  What  is  the  pressure  in  the  bearings  due  to 
the  belt? 


334  BELT! 

Solution  : 


900 

EXAMPLE  2. 

Suppose  the  diameters  of  the  pulleys  are  increased  until  a 
belt  speed  of  3000  feet  per  minute  is  obtained.  What  will  then 
be  the  pressure  in  the  bearings  caused  by  the  belt  when  trans- 
mitting 60  horse-power  ? 

Solution  : 

p  =  3  X  33000  X  60  =  1980  ^ 

3000 

By  the  above  examples  it  is  conclusively  shown  what  a 
great  advantage  there  is  in  using  pulleys  so  large  in  di- 
ameter that  proper  belt  speed  is  obtained.  (See  velocity  of 
belts,  page  337). 

The  approximate  pressure  may  also  be  very  conveniently 
obtained  from  the  width  of  the  belt,  thus:  For  light  single 
belts,  allow  1000  feet  of  belt  speed  per  horse-power  transmitted 
per  inch  of  width  of  belt.  The  effective  pull  in  such  a  belt  will 
be  33  pounds  per  inch  of  width,  and  the  pressure  on  the  bearings 
due  to  the  belt  will  accordingly  be  33  X  3  =  99  pounds  per  inch 
of  width  of  belt.  For  convenience,  say  100  pounds  pressure  in 
the  bearings  per  inch  of  width  of  such  belts.  For  belts  where 
800  running  feet  are  allowed  per  horse-power  per  inch  of  width 
of  belt,  this  reasoning  will  give  a  pressure  on  the  bearing  equal 
to  i23#  pounds  per  inch  of  belt.  For  convenience,  say  125 
pounds  pressure  in  the  bearings  per  inch  of  width  of  such 
belts.  For  belts  where  600  running  feet  are  allowed  per  horse- 
power per  inch  of  width,  the  pressure  in  the  bearing  is  equal 
to  165  pounds  per  inch  of  width  of  belt,  and  where  the  belt  is 
so  heavy  that  only  500  feet  of  belt  speed  per  horse-power  per 
inch  of  width  is  allowed,  the  pressure  in  the  bearings  will  be 
198  pounds  per  inch  of  width.  A  good,  practical  rule,  which 
can  very  easily  be  remembered,  is,  (when  belts  are  in  good  order 
and  have  the  proper  size  and  the  proper  tension)  : 

Multiply  weight  of  belt  in  ounces  per  square  foot  by  eight 
times'  the  width  of  the  belt  in  inches,  and  the  product  is 
approximately  the  pressure  in  pounds  upon  the  bearings  caused 
by  the  belt. 

EXAMPLE. 

A  belt  is  calculated  with  regard  to  the  horse-power  it  has 
to  transmit  under  a  given  velocity,  and  found  to  be  8-inch 
double  belting,  weighing  25  ounces  per  square  foot.  What  pres- 
sure will  it  cause  on  the  bearings  when  working  at  proper 
tension  ? 


BELTS.  335 

Solution,  by  the  last  rule : 


—  1600  pounds. 


Solution,  by  the  first  rule: 


At  a  speed  of  3000  feet  per  minute  such  a  belt  will  transmit 

X  8 
X)         : 

formula : 


~~500 — '  —  ^  horse-power,  and  calculating  the  pressure  by  the 


p  _  3  X  33000  X  H 


p  =  3  X  33000  X  48  =  15g4          ^ 
3000 

Both  rules  give  nearly  the  same  result,  and  one  is  just  as 
correct  as  the  other,  as  all  such  figuring  is  nothing  more  than 
approximation  at  the  best.  The  pressure  on  the  bearings  may 
be  a  great  deal  more  than  calculated  above.  Sometimes  the 
pulleys  are  roughly  made,  belts  are  poor,  and  consequently  the 
coefficient  of  friction  between  belt  and  pulley  is  small,  and  as 
the  belt  has  to  be  a  great  deal  tighter  in  order  to  do  the  work, 
the  pressure  on  the  bearing  will  be  greatly  increased.  Very 
frequently,  from  pure  ignorance  or  carelessness,  belts  sare  made 
very  much  tighter  than  necessary,  and  enormous  sums  of  money 
may  be  wasted  in  this  way  in  large  factories,  as  the  steam 
engines,  at  the  expense  of  the  coal  pile,  have  to  furnish  power 
not  only  to  do  the  useful  work,  but  also  to  overcome  all  the 
friction  produced  by  such  over-strained  belts,  hot  bearings,  etc. 
A  belt  will  transmit  more  power  over  a  good,  smooth  pulley 
than  over  a  rough  one.  When  pulleys  are  covered  with  leather 
a  belt  will  transmit  about  25%  more  power  than  it  will  when 
running  over  bare  iron  pulleys,  and  in  transmitting  the  same 
power  a  much  slacker  belt  may  be  used,  thereby  reducing  the 
friction  in  the  bearings. 


Special  Arrangement  of  Belts. 

By  the  use  of  suitable  guide  pulleys  it  is  possible  to  connect 
with  belts  shafts  at  almost  any  angle  to  each  other.  But 
experience  is  required  and  care  must  be  exercised  to  do  it  suc- 
cessfully. When  guide  pulleys  are  used  in  order  to  change  the 
direction  of  a  belt,  always  remember  that  when  the  belt  is  run- 
ning the  most  pressure  is  thrown  on  the  pulley  guiding  the 
working  part  of  the  belt.  This  pulley  is,  therefore,  very  liable 
to  heat  in  its  bearings,  if  not  designed  to  have  bearing  surface 
enough  and  also  to  have  proper  means  for  oiling. 


336 


BELTS. 


Fig.  2  shows  an  arrangement 
by  which  the  direction  of  motion 
of  two  shafts  may  be  reversed, 
when  the  distance  between  the 
shafts  is  too  short  for  the  use  of  a 
crossed  belt  or  when  a  crossed 
belt,  for  any  other  reason,  cannot 
be  used. 

Suppose  pulley  A  to  be  the 
driver  and  to  run  in  the  direction  of  the  arrow.  C  and  D  are 
guide  pulleys,  and  the  motion  of  the  driven  shaft  B  is  in  the  op- 
posite direction  to  the  shaft  A.  In  this  case  the  guide  pulley  C  is 
on  the  working  part  of  the  belt  and  is  the  one  to  which^special 
attention  must  be  paid  in  regard  to  heating.  If  the  direction 
of  shaft  A  is  reversed,  guide  pulley  D  will  be  on  the  working 
part  of  the  belt. 


Crossed  Belts. 

If  the  distance  between  A  and  B  (Fig.  2)  had  been  long 
enough,  it  would  have  been  preferable  to  reverse  the  motion  of 
B  by  means  of  a  crossed  belt,  instead  of  by  the  arrangement 
shown  in  Fig.  2. 

Crossed  belts  do  not  work  well  when  running  on  pulleys 
small  in  diameter  as  compared  to  the  width  of  the  belt. 

Too  short  distance  between  the  shafts  must  be  avoided. 

Wide  crossed  belts  are  very  unsatisfactory;  therefore, 
instead  of  running  one  wide  crossed  belt  it  is  preferable  to  use 
two  belts,  each  of  half  the  width,  and  run  them  on  two  separate 
pairs  of  pulleys.  Such  belts  should  be  of  equal  thickness,  and 
the  pulleys  should  be  crowned,  well  finished  and  of  correct 
size,  so  that  each  belt  will  do  its  share  of  the  work. 


Quarter =Turn  Belts. 

Fig.  3  shows  a  so-called  quarter- 
turn  belt,  used  to  connect  two  shafts 
when  running  at  an  angle  and  laying 
in  different  planes.  The  principal 
point  to  look  out  for  is  to  place  the 
pulleys  (as  shown  in  Fig.  3)  so  that 
the  belt  runs  straight  from  the  de- 
livering to  the  receiving  side  of  each 
pulley. 

The  pulleys  shown  in  Fig.  3  are 
set  right  for  belts  running  in  the 
direction  of  the  arrows.  If  the  mo- 
tion is  reversed,  the  belt  will  run  off 
the  pulleys. 


FIG.  3. 


BELTS.  337 

Angle  Belts. 

The  belt  arrangement  shown  in  Fig.  4  is  usually  called  an 
angle  belt,  and  is  used  to  connect  two  shafts  at  an  angle.    Either 
one,  A  or  B,  may  be  the  driver,  and 
there  are  two  guide  pulleys  (one  for  F|G-  4-     , 

each  part  of  the  belt  at  C),  one  of  which, 
of  course,  is  on  the  driving  part  of  the 
belt. 

Crossed  belts,  quarter-turn  belts, 
and  angle  belts  must  never  be  wide  and 
thin  ;  much  better  results  are  obtained 
by  narrow,  double  belts  than  by  wide, 
single  ones. 

Angle  belts  and  quarter-turn  belts 
are  frequently  bothersome  contrivances. 
Their  running  is  sometimes  improved 
by  making  a  twist  in  the  belt  when 
joining  its  ends ;  that  is,  lacing  the  flesh  side  of  one  end  and  the 
hair  side  of  the  other  end  on  the  outside.  This  will  prevent  one 
side  of  the  belt  from  stretching  more  than  the  other. 

Slipping  of  Belts. 

Owing  to  the  elasticity  of  belts,  there  must  always  be  more 
or  less  slip  or  "  creep "  of  the  belts  on  the  pulleys.  Under 
favorable  conditions  it  may  be  as  low  as  2%,  but  frequently  the 
slip  is  more.  Therefore,  if  two  shafts  are  connected  by  belts, 
and  both  should  have  very  nearly  the  same  speed,  the  diameter 
of  the  driver  should  be  at  least  2%  larger  than  the  diameter  of 
the  driven  pulley.  When  the  driver  is  comparatively  large  in 
diameter  and  the  driven  pulley  is  small,  it  is  advisable  to  have 
the  driver  from  2  to  5%  over  size,  in  order  to  get  the  required 
speed. 

Tighteners  on  Belts. 

If  tighteners  are  used  they  should  always  be  placed  on  the 
slack  part  of  the  belt. 

Velocity  of  Belts. 

Belts  are  run  at  almost  all  velocities  from  less  than  500  to 
5000  feet  per  minute,  but  good  practice  indicates  that  whenever 
possible  main  belts  having  to  transmit  quantities  of  power  are 
run  most  economically  at  a  speed  of  3000  to  4000  feet  per  min- 
ute. At  a  higher  speed  both  practice  and  theory  seem  to  agree 
that  the  loss  due  to  the  action  of  the  centrifugal  force  in  the  belt 
when  passing  around  the  pulley,  and  that  the  wear  and  tear  is  so 
great  when  the  speed  is  much  over  4000  feet  per  minute  that 
there  is  not  much  practical  gain  in  increasing  the  speed.  But, 
as  a  general  rule,  whenever  possible  the  higher  the  belt  speed 
the  more  economical  is  the  transmission  as  long  as  the  belt 
speed  does  not  exceed  the  neighborhood  of  4000  feet  per  minute. 


338  BELTS. 

Oiling  of  Belts. 

Belts  should  be  kept  soft  and  pliable  and  are,  therefore, 
usually  oiled  with  either  neat's-foot  oil  or  castor  oil.  Too  much 
piling  is  hurtful,  but  the  right  amount  of  oiling  at  proper  times 
is  very  beneficial  to  the  action  of  the  belt  and  will  prolong  its 
utility  to  a  great  extent. 

REMARKS. — All  previous  rules  for  calculating  belting  are 
founded  upon  good,  legitimate  practice,  but  are  only  offered  as 
a  guide,  as  no  rule  can  be  given  which  will  fit  all  cases. 

For  instance,  a  belt  may  be  amply  large  to  transmit 
a  given  horse-power  when  running  in  a  horizontal  direction, 
but  it  may  fail  to  do  the  same  work  if  running  in  a  vertical  di- 
rection. A  belt  may  be  large  enough  to  do  its  work  when  run- 
ning in  a  vertical  direction  over  pulleys  of  unequal  size  with 
the  large  pulley  on  the  lower  shaft,  but  it  may  fail  to  do  the 
same  work  satisfactorily  with  the  large  pulley  on  the  upper  shaft 
and  the  small  pulley  on  the  lower  one. 

Leather  belts  should  not  be  used  where  it  is  damp  or  wet, 
but  rubber  belting  will  usually  give  good  service  in  such 
places. 

For  information  regarding  rubber  belts,  see  manufacturers' 
catalogues. 


WIRE  ROPE  TRANSMISSION. 

Transmitting  power  by  wire  ropes  running  at  a  high  speed 
over  grooved  pulleys,  or  "  telodynamic  transmission."  as  it  is 
also  called,  is  the  invention  of  the  brothers  Him  of  Switzerland. 
For  long  distances  this  mode  of  transmission  is  far  cheaper  than 
leather  baiting  or  lines  of  shafting.  Fig.  1  shows  a  section  of  a 
pulley  as  u^ed  for  this  kind  of  transmission  ;  a  is  an  elastic  fill- 
ing, usually  made  from  leather  cut  out  and  packed  in  edgewise. 
The  groove  is  made  wide,  so  that  the  rope  will  rest  entirely 
against  the  packing  and  not  touch  the  iron.  This  is  different 
from  transmission  with  hemp  rope,  which  is  made  to  wedge  into 
the  groove  of  the  pulley. 

The  diameter  of  the  pulley  in  the  groove,  where  the  wire 
rope  runs,  ought  to  be  at  least  150  times  the  diameter  of  the 
rope  ;  the  larger  the  better,  so  long  as  the  velocity  of  the  rope 
does  not  exceed  5000  feet  per  minute.  The  pulleys  must  run 
true  and  be  in  balance  and  in  exact  line  with  each  other,  and  the 


WIRE  ROPE  TRANSMISSION. 


339 


shafts  must  be  parallel.  The  distance  between  shafts  should 
never  be  less  than  60  feet  and  should  preferably  be  from  150  lo 
400  feet.*  For  distances  longer  than  400  feet,  either  carrying 
pulleys  or  intermediate  jack  shafts  are  generally  used,  although 
spans  as  long  as  (500  feet  or  more  have  been  used,  but  only  when 
it  is  possible  to  give  the  rope  the  proper  deflection  without  its 
touching  the  ground.  Usually  the  speed  is  from  3000  to  6000 
feet  per  minute.  Higher  speed  would  be  dangerous  from  the 
stress  in  cast-iron  wheels  due  to  centrifugal  force. 


flj  g 

1& 


d  ins. 


b   ins. 


i5 


fi 
"f 

1 

u 

u 


H 


/"ins. 


I 

H 


H 
H 


S 


*t 

I 

« 


g  ins. 


H 
M 
H 


t  ins. 


A 

!i 


FIG.  i. 


Tightening  pulleys  should  not  be  used,  because  if  the  distance 
between  centers  of  shaft  is  top  short  to  give  the  proper  tightness 
to  the  rope  without  a  tightening  pulley,  wire  rope  transmission 
is  not  the  form  best  adapted  to  the  circumstances.  Guide 
pulleys  or  idlers  should  be  avoided  as  much  as  possible,  but 
when  necessary  they  should  be  as  carefully  made  and  put  up  as 
the  main  pulleys,  and  they  ought  not  to  be  less  than  half  the 
diameter  of  the  main  pulley  if  on  the  slack  part,  but  of  the  same 
size  if  they  are  on  the  tight  part  of  the  rope.  Wire  rope  for 
transmission  is  usually  made  from  the  best  quality  of  iron, 
has  seven  wires  to  a  strand  and  consists  of  six  strands  laid 
around  a  hemp  core  in  the  center  of  the  rope.  The  diameter 
of  the  wire  rope  is  from  nine  to  ten  times  the  diameter  of  each 
single  wire. 

Never  use  galvanized  rope  for  power  transmission,  but  pre- 
serve the  rope  by  painting  with  heavy  coats  of  linseed  oil  and 
lampblack. 


*  When  distance  between  shafts  is  less  than  60  feet,  leather  belts  are  prefer- 
able to  wire  rope. 


340  WIRE  ROPE  TRANSMISSION. 

Transmission  Capacity  of  Wire  Ropes. 

A  one-inch  rope  running  5000  feet  per  minute  is  capable  of 
transmitting  200  horse-power.  The  transmitting  capacity  of  the 
rope  is  in  proportion  to  the  square  of  its  diameter,  and  the  power 
transmitted  by  the  rope  when  the  velocity  is  less  than  5000  feet 
per  minute  is  practically  in  proportion  to  its  velocity.*  Hence 
the  formula: 

H=  d*  X  FX  20°  which  reduces  to  H  =  0.04  X  d*  X  V 

5000 

H  =  Horse-power  transmitted. 
d—  Diameter  of  rope  in  inches. 
V  —  Velocity  of  rope  in  feet  per  minute. 

EXAMPLE. 

How  many  horse-power  may  be  transmitted  by  a  wire  rope 
y2  inch  in  diameter  running  over  proper  pulleys  at  a  velocity  of 
2500  feet  per  minute  ? 

Solution : 

H=  0.04  X  l/2  X  y2  X  2500  =  25  horse-power. 
The  pressure  on  the  bearings  will  not  be  less  than  three 
times  the  force  transmitted,  and  may  be  calculated  thus  : 

Pressure  on  bearings  =  S  X  "Q^e-power  X  33000 
Velocity  in  feet  per  mm. 

EXAMPLE. 

What  will  be  the  least  pressure  in  bearings  for  a  wire  rope 
transmitting  150  horse-power  at  a  velocity  of  5000  feet  per 
minute  ? 

Pressure  on  bearings  =  2Jli50_><_??_020  =  2970  pounds. 

5000 

If  there  is  one  bearing  on  each  side  at  an  equal  distance 
from  the  pulley,  the  pressure  on  each  bearing  will  be  -2-y-ft  = 
1485  pounds.  This  is  the  calculated  pressure,  and  represents 
what  the  pressure  should  be,  but  it  is  not  certain  that  this  is  the 
actual  pressure.  It  may  be  greatly  increased  by  having  the  rope 
too  tight,  t 

*  When  the  velocity  of  the  rope  exceeds  6000  feet  per  minute  the  stress  caused 
by  centrifugal  force  when  the  rope  is  bending  around  the  pulley  considerably 
reduces  its  transmitting  capacity.  This  loss  increases  very  fast  above  this  speed, 
because  the  centrifugal  force  increases  as  the  square  of  the  velocity.  It  is  very 
doubtful  if  there  is  practically  any  gain  to  run  wire  ropes  at  a  speed  exceeding  6000 
feet  per  minute  when  wear  and  tear,  loss  due  to  centrifugal  force,  etc.,  are 
considered. 

t  Sometimes  a  pulley  is  put  on  the  free  end  of  a  line  of  shafting  projecting 
through  the  wall  and  drawn  by  a  wire  rope  outside  the  shop;  this  will  do  only 
when  3  comparatively  small  amount  of  power  is  to  be  transmitted. 


WIR£    ROPK    TRANSMISSION. 


341 


The  tension  of  the  rope  may  be  calculated  from  its  de- 
flection when  at  rest  (see  Fig.  2),  and  for  a  rope  running  hor- 
izontally the  usual  formula  is : 


(very  nearly) 


P  =  Force  in  pounds  at  f. 

W  =  Weight  of  rope  in  pounds  from  d  to  f,  which  is  half 
the  span. 

b  =  Half  the  span  in  feet. 

a  =  Twice  the  deflection  in  feet. 

NOTE.— (See  Fig.  2.)  If  the  length  of  the  line  a  represents 
the  weight  of  the  part  of  the  rope  from  */to/~,  the  length  of  the 
line  .r  represents  the  tension  in  the  rope  at/";  therefore  the  ten- 
sion will  be  as  many  times  the  weight  as  the  length  of  line  a  is 
contained  in  the  line  x. 


EXAMPLE. 

The  horizontal  distance  between  two  pulleys  is  200  feet ; 
when  standing  still  the  deflection  in  a  wire  rope  of  %" 
diameter  is  5  feet.  What  is  the  tension  in  the  rope  ? 

Solution : 

In  Table  No.  38  the  weight  of  %"  wire  rope  is  given  as  1.12 
pounds  per  foot;  therefore,  100  feet  of  %"  rope  will  weigh  112 
pounds. 

112  X  V  1002  +  10*     _     112 
10 


=  1125.6  pounds. 


This  is  the  tension  in  each  part  of  the  rope  ;  therefore  the 
force  against  the  pulley,  due  to  the  weight  of  the  rope,  is  1125.6 
X  2  —  2251.2  pounds.  If  this  is  supported  by  a  bearing  on  each 
side  of  the  pulley,  the  pressure  on  each  bearing,  if  both  are  the 
same  distance  from  the  pulley,  will  be  1125.0  pounds. 


342  WIRE  ROPE  TRANSMISSION. 

The  tension  is  increased  by  reducing  the  deflection.  For 
instance,  if  the  deflection  is  reduced  to  4  feet  the  tension  on  the 
rope  will  be, 

P  =  U2  X 


p=  112JOOO.S  =  1404.2  pounds. 

Thus,  the  tension  might  be  increased  to  any  amount  within 
the  ultimate  breaking  strength  of  the  rope. 

Deflection  in  Wire  Ropes. 

When  the  rope  is  in  motion  the  deflection  will  increase  on 
the  slack  side  and  decrease  on  the  tight  side  ;  therefore,  if  the 
span  is  long  the  rope  may  touch  the  ground  when  running  if  the 
pulleys  are  not  placed  on  sufficiently  high  towers.  There  is 
really  nothing  else  which,  within  practical  limits,  determines  the 
length  of  the  span,  which  may  just  as  well  be  1000  feet,  or  even 
more,  providing  the  proper  deflection  can  be  given  to  the  rope 
without  touching  the  ground.  When  possible  the  lower  part  of 
the  rope  should  be  the  working  side,  but  in  a  long  span  this  is 
impossible,  because,  when  running,  the  lower  part  of  the  rope 
would  be  tight  and  the  upper  part  slack,  causing  the  two  parts 
of  the  rope  to  strike  together,  which  must  never  be  allowed. 
When  the  length  of  the  span  exceeds  35  times  the  diameter  of 
the  pulleys  it  is  safest  to  have  the  upper  part  of  the  rope  the 
working  side  and  the  lower  part  the  slack  side. 

When  the  lower  part  of  the  rope  is  the  slack  side,  the  least 
space  allowable  for  the  slack  of  the  rope  at  the  center  of  the 
span  will  (when  the  rope  is  as  tight  as  given  in  Table  No.  38),  be 
obtained  by  the  formula  : 

Distance  =  0.00015  X  (span)2 

but,  to  allow  for  contingencies,  it  is  better  to  have  more  room. 
When  the  lower  side  of  the  rope  is  the  tight  side,  the  rope  will 
be  clear  from  the  ground  when  running  if  the  space  is  0.0001  X 
(span)'2.  The  deflection  in  the  rope  when  standing  still  which 
will  produce  a  pressure  on  the  bearings  and  give  tension  enough 
to  transmit  the  horse-power  given  in  Table  No.  38,  may  be  cal- 
culated approximately  by  the  formula  : 

d  —  0.00009  X  /2 

d  •=•  Deflection  in  feet. 

/  =  Distance  between  pulleys  in  feet.    (See  Fig.  2). 

EXAMPLE. 

The  distance  between  the  pulleys  being  400  feet,  find  the 
greatest  allowable  deflection  in  the  rope,  when  standing  still,  in 
order  to  transmit  the  horse-power  given  in  Table  No.  38. 


WIRE   ROPE   TRANSMISSION. 


343 


Solution : 

d  =  0.00009  X  400  X  400  —  14.4  feet. 

When  the  rope  is  new  it  is  always  put  on  with  more 
tension  than  is  necessary  to  transmit  the  power,  because  new 
rope  will  stretch.  It  is,  therefore,  very  important  when  de- 
signing such  transmission  to  calculate  the  maximum  pressure 
which  the  rope  will  exert  on  the  bearings  when  put  on  with  the 
least  deflection  ever  wanted,  and  calculate  size  of  bearings  and 
shafting  for  pulleys  according  to  this  stress,  with  due  considera- 
tion not  only  for  strength  but  also  for  heat  and  wear.  (See  page 
360  and  page  367.)  The  correct  amount  to  allow  for  stretch  will 
vary  with  different  kinds  of  rope  and  also  with  the  tempera- 
ture. If  a  rope  is  spliced  on  a  warm  summer  day  it  must  be 
made  slacker  than  if  it  was  spliced  on  a  cold  winter  day,  as  the 
length  of  the  rope  will  be  changed  considerably  by  the  difference 
in  temperature  ;  the  only  guide  is  practical  experience  and  good 
judgment.  As  a  general  rule,  it  may  be  safe  to  allow  about  half 
of  the  deflection  as  previously  calculated  when  splicing  a  new 
rope,  provided  that  the  shafts  and  bearings  are  constructed  so 
as  to  allow  such  tension.  The  rope  is  always  strong  enough. 
The  splicing  of  the  rope  should  be  done  by  a  man  ex- 
perienced in  that  kind  of  work.  The  splice  itself  is  usually 
made  at  least  240  times  the  diameter  of  the  rope. 

TABLE  No.  38. -Giving  Suitable-Sized  Pulleys  for  Different  Sizes  of 
Wire  Rope,  Weight  of  Rope,  Horse-Power  which  Different  Sizes 
of  Wire  Rope  Hay  Transmit  at  Different  Velocities,  the  least  Stress 
at  which  it  may  be  done  and  the  Least  Corresponding  Pressure  on 
the  Bearings  ;  also,  the  Ultimate  Average  Strength  of  Wire  Rope. 


1 

i 

i 

1 

i 

•S 

0 

Horse-Power  Trans- 
mitted at  Different 

1 

| 

ll 

1  1 

1 

O) 

H 

"c  « 

1  1 

Velocities. 

Jj 

3 

II 

§  § 

!| 

«J 

^o 

«  M 

1*6 

V 
3 

1 

3 

3 

.s  s 

w-S  w 

*C  r^ 

cP^ 

42   fi 

•c  g 

C 

.S 

.3 

.S 

jj 

Is 

1  ll 

V 

|| 

I'D 

o  S 

s| 

Is 

s' 

b 

fcl 

*o 

S* 

H^-^ 

1 

"*  w 

'5  8, 

0   | 

S, 

8, 

i 

i 

1 

f£ 

f  1 

1 

g 

d 
.2 

ll 

go 

1 

1 

1 

1 

£ 

1 

tj}  ^ 

.§      c 

*S5 

c 

tfl  _c 

A 

.2 

.2 

V 

a     £ 

'C 

s 

s 

P-4 

~ 

•~  ** 

o 

o 

$ 

Q 

Q 

^ 

D     C, 

O 

H 

H 

OH 

« 

^ 

S 

5 

y 

0.21 

4500 

1S7 

187 

374 

285.5    501 

14 

17 

22 

28 

6 

iV 

0.23 

6000 

253 

253 

506 

379.5 

759 

19 

23 

31 

38 

7 

0.31 

8000 

330 

330 

660 

495 

990 

25 

30 

40 

50 

8 

fs 

0.57 

12000 

517 

517 

1034 

775.5 

1551 

39 

47 

62 

78 

10 

& 

0.92  18000    636!  6361272 

954 

1908 

56 

67 

90 

112 

12 

1.12 

24000  1012  1012  2024 

1518 

3036 

77 

92 

122 

153 

14 

1 

1.50 

320001320,13201264011985 

3960 

100 

120 

160 

200 

344  wlRE    ROPE    TRANSMISSION. 

EXAMPLE. 

From  a  shaft  running  150  revolutions  per  minute  100  horse- 
power is  to  be  taken  off  by  a  wire  rope.  The  velocity  of  the  rope 
is  to  be  5000  feet  per  minute.  What  size  of  pulley  and  rope 
will  be  required  ? 

Solution : 

In  Table  No.  38  it  will  be  found  that  a  ^-inch  wire  rope,  run- 
ning at  5000  feet  per  minute,  is  capable  of  transmitting  112  horse- 
power ;  thus,  select  a  %-inch  rope.  The  diameter  of  the  pulley 

will  be  i5Q  x  3  1416  =  10'6  feetp  In  the  table  'lt  wil1  be 
found  that  a  10-foot  pulley  is  the  smallest  advisable  to  run 
with  a  %-inch  wire  rope,  therefore  the  pulley  10.6  feet  in  diame- 
ter is  within  the  requirements.  The  next  step  is  to  calculate  the 
pressure  on  the  bearings.  In  the  table  it  is  found  that  the  least 
pressure  due  to  the  transmission  of  112  horse-power  is  1908 
pounds.  This  cannot  be  used  in  calculating  sizes  of  shafts  and 
bearings,  but  use  the  maximum  pressure,  which  is  calculated  ac- 
cording to  the  allowable  deflection  in  the  rope,  as  explained 
on  page  341.  Also  consider  weight  of  pulley  and  shaft,  then 
calculate  size  of  shaft  and  bearings,  with  due  consideration  to 
strength,  stiffness,  wear,  heat,  etc.  (  See  pages  360-367.) 

Transmission  of  Power  by  Manila  Ropes. 

Manila  ropes  are  used  more  or  less  for  transmission  of 
power.  In  this  country  one  continuous  rope,  going  back  and 
forth  in  separate  grooves  over  the  pulleys  several  times,  is  fre- 
quently used,  and  a  tightening  arrangement  is  placed  on  one  of 
the  slack  parts,  which  automatically  keeps  the  rope  at  the 
proper  tension,  regardless  of  changes  due  to  weather  or  stretch 
due  to  wear.  This  arrangement  has  its  advantages  in  keeping 
the  rope  at  more  even  tension  than  is  possible  with  the  Euro- 
pean system,  but  the  disadvantage  is  that  if  a  break  occurs  the 
transmission  is  entirely  disabled  until  it  is  repaired.  The  Euro- 
pean practice  is  to  use  several  single  ropes  running  in  separate 
grooves  side  by  side  on  the  same  pulley.  This  has  the  advan- 
tage that  if  one  of  the  ropes  should  break  it  is  usually  possible 
to  run  undisturbed  until  there  is  a  chance  to  repair  it,  because 
it  is  always  advisable  to  have  margin  enough  in  the  transmission 
capacity  of  the  ropes  so  that  the  shaft  will  run  satisfactorily, 
even  if  one  rope  is  taken  off.  The  disadvantage  of  this  system 
is  the  difficulty  in  keeping  all  the  ropes  at  equal  tightness  and 
getting  them  to  pull  evenly. 

Fig.  3  shows  the  usual  shape  of  pulley  used  for  manila 
ropes,  which  may  be  made  from  either  wood  or  iron.  The 
European  practice  is  to  use  iron,  but  whichever  material  is  used 
it  is  very  important  to  have  the  sides  of  the  grooves  care- 
fully polished,  as  the  rope  rubs  on  the  sides  in  entering  and 


MANILA    ROPE    TRANSMISSION. 


345 


leaving  the  pulley  and  will  wear  out  in  a  short  time  if  the 
pulley  is  left  as  it  comes  from  the  lathe  tool.  Sand  and  blow- 
holes must  also  be  avoided.  The  angle  of  groove  is  usually  45°, 
and  the  rope  is  made  to  wedge  into  it,  as  shown  in  Fig.  3. 

The  usual  shape  of  grooves  for  guide  pulleys  is  shown  in 
Fig.  4. 


rt  (D 

r>  cx 


</ins.    a  ins.1  <r  i.is. 


1 

H 
if 

2 


If        t 

Hi    I 


1 


M 


X 


A 

I 

TV 


s 


^•ins. 


I 


1 

H 

H 

2 

3* 

4^ 


FIG.  4. 


The  best  speed  for  ropes  is  from  1500  to  5000  feet  per 
minute.  When  the  velocity  of  the  rope  exceeds  6000  feet  per 
minute  the  loss,  due  to  the  centrifugal  force,  is  so  great  that  it 
will  hardly  pay  to  increase  the  velocity.  The  diameter  of  the 
pulleys  ought  to  be  at  least  50  times  the  diameter  of  the  rope. 


Transmission  Capacity  of  Manila  Rope. 

A  manila  rope  two  inches  in  diameter,  running  over 
properly-shaped  pulleys  at  a  speed  of  5000  feet  per  minute,  is 
capable  of  transmitting  50  horse-power.  The  transmitting 
capacity  of  the  rope  is  in  proportion  to  the  square  of  its 
diameter,  and  the  power  transmitted  by  the  rope  is  in  propor- 
tion to  its  velocity ;  therefore,  a  one-inch  rope,  running  5000  feet 
per  minute,  will  transmit  12%  horse-power,  and  the  formula 
will  be : 

Horse-power  =   *  X  *  *  12'5 

5000 
which  reduces  to 

Horse-power  =  0.0025  X  d*  X  v 
d  =  Diameter  of  rope  in  inches. 
v  —  Velocity  of  rope  in  feet  per  minute. 


346 


MANILA   ROPE   TRANSMISSION. 


EXAMPLE. 

What  horse-power  may  be  transmitted  by  a  manila  rope 
\Yz  inches  in  diameter,  running  over  nine-foot  pulleys  at  a  speed 
of  150  revolutions  per  minute  ? 

Solution : 

Nine-foot  pulleys,  running  150  revolutions  per  minute,  give 
the  rope  a  velocity  of  3.1416  X  9  X  150  =  4241  feet  per  minute, 
and  the  horse-power  transmitted  will  be : 

H-P  =  0.0025  X  ll/2  X  ll/2  X  4241 

H-P  =  0.0025  X  2#  X  4241 

H-P  =  23.85 ;  practically,  24  horse-power. 

Weight  of  Manila  Rope. 

The  weight  of  one  foot  of  manila  rope  of  one-inch  diameter  is 
T3o  pound ;  therefore,  the  weight  per  foot  of  any  size  may  be 
calculated  approximately  by  the  formula : 

W=d*  X  0.3 
dT—  Diameter  of  rope  in  inches. 

W  •=•  Weight  of  rope  in  pounds  per  foot. 

EXAMPLE. 

What  is  the  weight  of  360  feet  of  manila  rope  of  1^-inch 
diameter? 

Solution : 

Weight  of  360  feet  =  0.3  X  360  X  ll/2  X  1%  =  243  pounds. 

TABLE  No.  39, 

Giving  the  Weight  of  Rope  in  Pounds  per  Foot,  Driving  Force  in 
Pounds,  and  Corresponding  Horse=Po\ver  Transmitted  at  Differ* 
ent  Velocities. 


*l 

d 

l| 

1 

Horse-power  Transmitted  at  Different  Velocities. 

Ij 

0  £ 

jjl 

. 

. 

. 

. 

if 

"o.S 

^p. 

3 

3 

1 

3 

3 

3 

3 

3 

p^ 

i 

— 

bo 

«1 

11 

u.S 

li 

1*1 

«.S 

IJ 

o3-S 

^ 

i 

bO 

fa    B 

fa   5 

fa  - 

fa   fa 

fa  b 

fa   ° 

fa    E 

fa  " 

J 

I 

Q 

1* 

1" 

§  ft 

1" 

1" 

I" 

1" 

i* 

2 

X 

0.075 

21 

0.94 

1.25 

1.56 

1.87 

2.18 

2.50 

3.12 

3.75 

2/4 

X 

0.12 

33 

1.45 

1.94 

2.24 

2.90 

3.39 

3.87 

4.48 

5.81 

2/2 

^  10.16 

47 

2.11 

2.81 

3.52 

4.82 

4.92!  5.62 

7.03 

8.44 

3/^ 

1      |0.80 

83 

3.75 

5 

6.25 

7.50 

8.7510 

12.50 

15 

4X 

IX  '0.47 

132 

5.86 

7.81 

9.77 

11.7213.7315.62 

19.23 

23.44 

5 

IX 

0.67 

186 

8.44 

11.25 

14.00 

16.87'l9.69  22.50 

28.12 

33.75 

6^ 

1& 

0.92 

255 

11.48 

15.31 

18.31 

22.97 

26.79  30.62 

36.61 

45.94 

8 

2 

1.20 

330 

15 

20 

25 

30      35      |40 

50 

60 

MANILA    ROPE    TRANSMISSION.  347 

The  transmitted  horse-power,  as  given  in  Table  No.  39,  is 
calculated  by  the  formula, 

H-P  =  0.00025  X  d*  X  v 

EXAMPLE.     (Showing  application  of  Table  No.  39.) 

What  size  of  rope  is  required  to  transmit  50  horse-power 
when  three  independent  ropes  are  used,  running  over  the  same 
pulley  at  a  velocity  of  4000  feet  per  minute  ? 

Solution  : 

It  is  always  advisable  to  select  ropes  having  sufficient  trans- 
mitting capacity  to  continue  the  transmission  undisturbed,  even 
if  one  rope  breaks ;  therefore,  select  ropes  of  such  size  that  two 
ropes  will  transmit  nearly  25  horse-power  each.  In  Table  No. 
39  it  is  found  that  a  manila  rope  ll/2  inches  in  diameter,  running 
4000  feet  per  minute,  will  transmit  22.5  horse-power.  Thus,  this 
will  be  the  size  of  rope  to  use.  The  small  pulley  in  the  trans- 
mission must  not  be  less  than  five  feet  in  diameter.  (See  Table 
No.  39.) 

The  pressure  on  the  bearings,  due  to  tension  of  the  rope, 
will  not  exceed  three  times  the  driving  force,  because  manila 
ropes  run  comparatively  slack,  as  the  adhesion  to  the  pulley 
does  not  depend  so  much  on  the  tightness  of  the  rope  as  it  does 
on  its  wedging  into  the  groove  in  the  pulley.  The  driving 
force  of  manila  rope  of  IX -inch  diameter  is  given  in  the  table 
as  132  pounds;  therefore,  the  pressure  due  to  one  rope  will  be 
3  X  132  =  396  pounds,  and  the  pressure  due  to  three  ropes  will 
be  3  X  396  =  1188  pounds ;  besides  this,  the  weight  of  the  shaft 
and  the  pulley  should  be  considered  when  calculating  the  size  of 
shaft  and  bearings,  with  due  consideration  for  strength,  stiffness, 
wear,  heat,  etc.  (See  page  367.) 


Preservation  of  Manila  Rope. 

The  life  of  the  rope  is  prolonged  by  slushing  once  in  a 
while  with  tallow  mixed  with  plumbago.  The  rope  will  not  only 
wear  on  the  outside  but  also  within  itself,  because  the  fibers 
chafe  on  each  other  as  the  rope  bends  over  the  pulleys ;  hence 
the  preference  for  pulleys  of  large  diameter.  If  the  rope  is  not 
specially  prepared  for  transmission  purposes,  it  ought  to  be 
soaked  in  a  mixture  of  plumbago  and  melted  tallow  when  new, 
before  it  is  used.  There  is  on  the  market  manila  rope  especial- 
ly manufactured  for  transmission  purposes,  having  the  fibers 
treated  with  plumbago  and  tallow,  and,  whenever  obtainable, 
such  rope  should  be  used,  as  it  will  last  much  longer  and  give 
much  better  service  than  ordinary  manila  rope. 


348 


PULLEYS. 


PULLEYS. 

The  following  empirical  rule  gives  arms  of  nice  shape  and 
good  proportions  : 

When  the  diameter  of  the  pulley  is  at  least  4  times  its  face, 
use  6  arms  for  pulleys  from  12  to  60  inches.  For  a  12-inch  pulley 
make  the  arms  1 X  inches  wide  at  the  hub  and  add  T^  of  an  inch 
to  the  width  of  the  arm  for  each  inch  the  pulley  is  increased  in 
diameter. 

Formula :  * D  —  12     i   i  y 

16 
h  =  Width  of  arm  in  inches  projected  to  center  of  hub. 

(See  Fig.  1.) 
D  =  Diameter  of  pulley  in  inches. 

EXAMPLE. 

Find  width  of  arms  at  the  hub  for  a  60-inch  pulley. 

Solution : 

60  —  12 
h  =  — ^g +  IX  —  4X  inches. 

The  width  of  the  arm  at  the  rim  should  be  three-fourths  of 
the  width  at  the  hub,  and  the  thickness  should  be  one-half  of  the 
width  for  arms  with  segmental  sections  (see^x  Fig.  1)  and  four- 
tenths  of  the  width  for  elliptical  form  of  section  ;  (see  x  Fig.  1.) 
For  double  belts  multiply  these  diameters  by  1.3. 

Large,  well-round- 
ed fillets  must  be  used 
where  the  rim  and  arms 
meet  at  #2.  (See  Fig.  1.) 
For  very  wide  pulleys 
it  is  always  better  to  use 
two  sets  of  arms.  For 
small  pulleys,  under  12 
inches  in  diameter,  4 
arms  are  better  than  0, 
as  they  are  less  liable  to 
break  while  being  cast. 
Using  4  arms,  the  width 
of  the  arm  at  h,  in  a 
pulley  10  inches  in  di- 
ameter, may  be  1T\  in. ; 
pulley  8  inches  in  di- 
ameter, 1 1/$  in. ;  pulley  6 
inches  in  diameter,  1 
inch  ;  and  the  thickness 
and  taper  as  given 
above.  When  pulley  arms  crack  from  shrinkage  in  casting,  the 
trouble  may  usually  'be  prevented  by  either  increasing  the  thick- 
ness of  the  rim  of  the  pattern  or  by  reducing  the  size  of  the  hub, 


FIG.  1 


m 


PULLEYS.  349 

or  both ;  it  will  also  help  the  matter  to  remove  the  core  and  the 
sand  from  the  hub  as  soon  as  possible  after  the  pulley  is  cast, 
and  leave  the  casting  in  the  sand  undisturbed  until  cool. 

When  the  diameter  of  the  shaft  is  less  than  4  inches,  the  di- 
ameter of  the  hub  is  usually  made  twice  the  diameter  of  the 
shaft.  When  shafts  are  over  4  inches  in  diameter  the  hub  of 
the  pulley  is  usually  made  a  little  less  than  twice  the  diameter 
of  the  shaft.  The  length  of  the  hub  may  be  made  three-fourths 
the  width  of  the  rim,  for  a  tight  pulley,  and  five-fourths  the  width 
of  the  rim  for  a  loose  pulley. 

The  thickness  of  rim,  measured  at  the  edge,  is  usually : 

For  pulleys  under  12  inches  in  diameter,  T3g  inch. 

For  pulleys  from  12  to  24  inches  in  diameter,  #  inch. 

For  pulleys  from  24  to  36  inches  in  diameter,  ^  inch. 

For  pulleys  from  36  to  48  inches  in  diameter,  A  inch. 

For  pulleys  from  48  to  60  inches  in  diameter,  y2  inch. 

For  double  belts  increase  the  thickness  of  the  rim  one- 
eighth  of  an  inch. 

The  thickness  in  the  middle  may  be  about  \%  times  the 
thickness  at  the  edge. 

Pulleys  which  are  to  run  at  high  velocity  ought  to  be  turned 
both  inside  and  outside,  in  order  to  be  in  good  balance.  Pulleys 
to  go  on  line  shafts  ought  to  be  made  in  halves,  so  that  they  can 
be  put  on  and  taken  off  the  shaft  with  convenience.  Pulleys  on 
which  the  belts  are  to  be  shifted  must  be  a  little  over  twice  as 
wide  as  the  belt,  and  they  should  be  turned  straight  across  the 
face  on  the  outside.  Pulleys  on  which  the  belts  are  not  to  be 
shifted  ought  to  be  only  1.2  times  as  wide  as  the  belt,  and  they 
ought  to  be  turned  curved  across  the  face;  that  is,  the  outside 
diameter  of  the  pulley  must  be  largest  at  the  middle.  Most  fre- 
quently a  straight  taper  is  turned  each  way  from  the  middle  to 
the  edges,  and  the  following  proportions  will  give  good  results  : 

Pulleys  under  six  inches  wide,  |^-inch  taper  per  foot. 

Pulleys  from  6  to  12  inches  wide,  J^-inch  taper  per  foot. 

Pulleys  from  12  to  18  inches  wide,  24-inch  taper  per  foot. 

When  pulleys  are  turned  in  a  lathe  where  the  tail-stock  can  be 
set  over,  a  taper  of  ^-inch  per  foot  is  practically  obtained  when 
the  tail-stock  is  set  over  j^-inch  per  1  inch  lengtn  of  arbor.  For 
instance,  if  a  crown  pulley  is  to  be  turned  %-inch  per  foot,  and 
the  arbor  is  12  inches  long,  the  back  center  must  be  set  over  \\ 
=  ^5-inch.  If  the  arbor  had  been  14  inches  long  the  back  cen- 
ter would  have  had  to  be  set  over  £f  =  ^-inch  to  obtain  the 
same  result. 

All  pulleys  must  be  well  rounded  on  the  edges.  They  must 
also  be  carefully  balanced,  especially  if  they  are  to  run  at  high 
speed. 

Loose  pulleys  ought  to  have  longer  hubs  than  tight  pulleys. 
They  ought  never  to  have  hubs  shorter  than  the  width  of  the, 
rim,  and  must  always  be  provided  with  means  for  oiling. 


350  PULLEYS. 

Stepped  Pulleys. 

Stepped  pulleys,  or  cone  pulleys,  as  they  are  usually  called, 
may  be  considered  as  several  pulleys  of  different  diameters  cast 
together.  Their  proportions  and  sizes  are  calculated  to  get  the 
required  changes  of  speed,  and  the  belt  must  have  practically 
the  same  tension  on  all  the  different  changes. 

Frequently  it  is  required  to  have  both  pulleys  of  the  same  size, 
in  order  that  they  may  be  cast  from  the  same  pattern.  In  such 
cases  the  shaft  of  constant  speed  (usually  a  counter-shaft)  must 
be  run  at  a  velocity  equal  to  the  square  root  of  the  product 
of  the  fastest  and  the  slowest  speed  of  the  shaft  of  changeable 
speed  (which  usually  is  a  spindle  in  a  lathe  or  a  similar  machine). 
For  convenience,  in  the  following  formulas  we  will  call  the 
driver,  which  is  the  shaft  of  constant  speed,  a  counter-shaft,  and 
the  shaft  of  changeable  speed,  a  spindle. 

The  number  of  revolutions  of  the  counter-shaft  per  minute 
is  calculated  by  the  formula  : 


N  =  Number    of    revolutions    of    the    counter-shaft   per 

minute. 
F  =  Number   of  revolutions  of  the  spindle  per  minute, 

when  run  at  its  fastest  speed. 
6"  =  Number  of  revolutions  of  the  spindle  per  minute,  when 

run  at  its  slowest  speed. 

The  diameter  of  either  the  largest  or  the  smallest  step  is 
then  obtained  by  choosing  one  diameter  and  calculating  the 
other  by  the  formula  : 

D  —  d  X  N  d  —    D  X  * 

~n  N 

D  =  Diameter  of  largest  step  on  spindle. 
d  =  Diameter  of  smallest  step  on  counter-shaft. 
n  =  Slowest  number  of  revolutions  of  the  spindle  per 

minute. 

N  =  Revolutions  of  the  counter-shaft  per  minute. 
The  intermediate  steps  may  be  obtained  by  drawing  a 
straight  line,  a  b,  and  constructing  steps  within  the  angle 
formed  by  the  line  a  b  and  the  center  line  (see  Fig.  2).  The  sum 
of  the  diameters  of  the  two  opposite  steps  will  then  be  equal, 
and  this  is  the  way  in  which  stepped  pulleys  may  primarily  be  laid 
out,  whether  both  pulleys  are  of  the  same  size  or  not.  After- 
wards the  diameters  will  have  to  be  slightly  changed,  in  order 
to  give  the  belt  the  same  tension  on  any  of  the  different  steps, 
as  explained  further  on. 

EXAMPLE  1. 

A  pair  of  stepped  pulleys,  for  four  changes  of  speed,  both 
pulleys  of  the  same  size,  are  to  be  used  on  a  milling  machine 
spindle  and  its  counter,  the  fastest  speed  to  be  250  revolutions, 


PULLEYS.  351 

and  the  slowest  speed  00  revolutions,  per  minute.  The  diam- 
eter of  the  largest  step  is  15  inches.  What  should  be  the  speed 
of  the  counter-shaft,  and  what  is  the  diameter  of  each  inter- 
mediate step  ? 

Solution  : 

Speed  of  counter  =  V'ltf)  X  250  =  150  revolutions  per 
minute. 

The  diameter  of  the  largest  step  is  15". 

90  X  15 
Diameter  of  smallest  step  =     —  ^  —  =  9  inches. 

By  the  method  as  shown  in  Fig.  2,  the  intermediate  diam- 
eters are  found  to  be  11"  and  13".  The  speed  of  spindle  will  be  : 

First  speed      =    15°  X  15   =  250  revolutions  per  minute. 
9 

150  X  13 
Second  speed  =  --  ^  --  =  177  revolutions  per  minute. 

Third  speed    =   15°      11  —  127  revolutions  per  minute. 
13 

Fourth  speed  =  -   '       *      =    90  revolutions  per  minute. 
15 

These  calculations  are  only  correct  for  speed,  and  must  be 
slightly  modified  in  order  to  get  the  proper  tension  on  the  belt, 
if  an  open  belt  is  used  ;  for  a  crossed  belt  the  tension  is  cor- 
rect if  the  pulleys  are  laid  out  in  this  manner.  (See  page  352.) 

When  the  number  of  revolutions  per  minute  for  each  change 
of  speed  is  given,  the  diameters  of  the  intermediate  steps  may, 
with  regard  to  speed,  be  calculated  by  the  following  formulas  : 


DI=  Diameter  of  any  step  on  spindle. 
D  •=•  Diameter  of  largest  step  on  spindle. 
d  =  Diameter  of  smallest  step  on  counter-shaft. 
n  •=-  Revolutions  of  spindle  per  minute,  corresponding  to 

the  diameter  D\. 

N  =  Revolutions  of  counter-shaft  per  minute. 
After  the  diameter  of  any  step  on  the  spindle  is  calculated, 
the  diameter  of  the  corresponding  step  on  the  counter-shaft  may 
be  obtained  by  subtracting  the  diameter  of  the  step  on  the  spin- 
dle from  the  value  of  (D  +  d). 

EXAMPLE  2. 

A  lathe  spindle  is  required  to  run  at  40,  120  and  360  revolu- 
tions per  minute,  and  the  diameter  of  the  largest  step  is  18 
inches.  Calculate  speed  of  counter-shaft  and  diameter  of  steps. 


352 


PULLEYS. 


Solution : 

Speed  of  counter-shaft  will  be: 

N  =  \/  F  X  -S1  =  V  360  X  40  =  120  revolutions  per  minute. 
Diameter  of  smallest  step  on  spindle  will  be : 


18  X  40 
120 


=  6  inches. 


Diameter  of  the  intermediate  step  on  spindle  will  be: 
Z>i  =  (18  +  6)_XJL20  _  12  inches< 

(120  +  120) 

Thus,  the  sizes  of  each  step,  with  regard  to  speed,  should 
be  6,  12  and  18  inches,  but  with  regard  to  belt  tension  these 
sizes  have  to  be  slightly  altered. 


To  Correct  the  Diameter  of  Stepped  Pulleys  so  that  the 
Belt  will  have  the  Same  Tension  on  all  the  Steps. 

At  first  thought,  it  may  seem  as  if  the  belt  would  have 
equal  tension  on  each  step  when  the  sum  of  the  diameters  of 
the  largest  and  the  smallest  steps  of  the  two  pulleys  are  equal 
to  the  sum  of  the  diameters  of  the  two  middle  steps ;  but  this  is 
only  correct  if  a  crossed  belt  is  used  on  the  pulleys.  For  a  two- 
step  pulley  it  is  also  correct  for  either  open  or  crossed  belt,  if 
both  pulleys  are  of  the  same  size  ;  but  if  the  pulleys  are  of  dif- 
ferent sizes,  the  diameter  of  the  steps  must  be  calculated  for 
two  steps  as  well  as  if  there  were  more. 

It  is  evident  from  Fig.  2,  that  an  open  belt  will  be  tighter 
over  the  largest  and  the  smallest  pulleys  than  it  would  be  over 
the  two  middle  pulleys,  as  the  part  a  of  the  belt  runs  parallel  to 
the  center  line  and  will  be  as  long  as  the  distance  between 
centers,  but  the  inclined  line,  b,  will  be  as  much  larger  as  the 
distance  dto  e.  (See  Fig.  2). 


A  convenient  way  to  solve  this  is  :  First  calculate  pulleys 
that  will  give  the  required  speed,  and  of  such  sizes  that  the  sum 
of  the  diameters  of  the  two  steps  which  are  to  work  together 
will  be  equal,  then  calculate  the  length  of  the  belt  when  laying 
on  the  largest  and  smallest  steps,  with  a  given  distance  between 


PULLEYS.  353 

the  centers  of  the  shafts.  Then,  by  calculating  the  same  way, 
try  the  belt  on  the  other  steps,  which  will  then  have  to  be  cor- 
rected until  the  belt  will  fit  each  of  the  different  pairs  of  steps. 

The  length  of  the  belt  can  be  most  conveniently  calculated  by 
the  geometrical  rule  that  the  square  of  the  perpendicular  added 
to  the  square  of  the  base  is  equal  to  the  square  of  the  hypothe- 
nuse.  (See  page  150.)  The  space  between  the  centers  of  the 
shafts  is  considered  as  the  base,  and  the  difference  in  radius  of 
the  two  corresponding  steps  is  considered  as  the  perpendicular, 
which  are  both  known,  and  from  this  the  length  of  the  line  b  is 
calculated  (see  Fig.  2),  which  is  considered  as  the  hypothenuse. 
Assuming  that  the  belt  covers  half  the  circumference  of  both 
pulleys,  the  length  of  the  belt  can  be  found  by  adding  half 
the  circumference  of  each  step  to  twice  the  length  of  b. 

NOTE. — This  mode  of  calculation  is  not  exactly  correct,  but 
is  very  well  within  practical  requirements. 

The  length  of  half  the  circumference  of  the  pulley  is  most 
conveniently  obtained  by  the  use  of  Table  No.  24,  page  209,  by 
dividing  the  circumference  of  the  corresponding  circle  by  2. 

A  practical  rule  is  simply  to  calculate  the  distance  from  d 
to  *,  and  for  each  Tyinch  the  belt  is  found  to  be  too  [long,  add 
aVinch  to  the  diameter  of  the  corresponding  step  on  each  pulley. 

For  instance,  the  stepped  pulleys  in  Example  No.  2  are  cal- 
culated so  that  they  will  give  the  required  speed  to  the  machinery 
when  the  three  steps  are  18,  12  and  6  inches  in  diameter  and 
both  pulleys  are  equal.  Assume  the  distance  between  centers 
to  be  5  feet.  What  will  be  the  diameter  of  the  middle  step,  after 
it  has  been  corrected  so  that  it  will  give  the  right  tension  to  the 
belt? 

Solution : 

Five  feet  =  60  inches,  and  the  difference  between  the 
radius  of  the  corresponding  steps  is  9  —  3  =  6  inches.  The 
distance  from  e  to  d  will  be  : 

x  —  \/602  +  62  —  60  =  V  3636  —  60  =  60.3  —  60  =  0.3 

Thus,  each  part  of  the  belt  will  be  0.3"  too  long,  or  the  whole 
belt  will  be  0.6"  too  long  when  on  the  middle  step  ;  therefore,  in 
order  to  make  up  for  this,  the  middle  step  on  each  pulley  must 
be  increased  /.,"  =  fY'  in  diameter.  Thus,  the  middle  step  on 
each  pulley  will  be  12T3ff  inches  instead  of  12  inches  in  diameter ; 
but,  as  both  pulleys  are  increased,  this  does  not  change  the 
relative  speed  of  the  shafts  when  the  belt  is  on  the  middle  step, 
and  the  similarity  of  the  pulleys  is  also  preserved,  which  will 
admit  that  both  may  be  cast  from  the  same  pattern. 

The  square  root  of  3636  may  be  obtained  by  use  of  log- 
arithms (see  page  71),  thus  : 

. log.  3636        3.560624 

Log.  X/3636  =    -^ =  —% =  1-780312 

and  the  number  corresponding  to  this  logarithm  is  60.3. 


354  PULLEYS. 

Stepped  Pulleys  for  Back = Geared  Lathes. 

On  machinery  having  changeable  reducing  gearing,  such  as 
lathes,  milling  machines,  etc.,  it  is  frequently  the  aim  of  the 
designer  to  arrange  the  speed  of  the  counter  and  the  diameters 
of  the  different  steps  of  the  cone  pulley  in  such  proportions  that 
the  same  ratio  of  speed  will  be  maintained  on  each  step  and  also 
from  the  slowest  speed,  with  back  gears  out,  to  the  fastest 
speed,  with  back  gears  in.  When  the  ratio  of  the  back  gearing 
is  given,  the  ratio  of  speed  for  each  step  will  be  obtained  by  the 
formula : 

m  

s  =  V^ 

S  =  Ratio  of  speed  for  each  step. 

m  =  Number  of  changes  of  speed  on  the  cone  pulley. 

7?  =  Reduction  of  speed  by  the  back  gearing. 

EXAMPLE. 

The  back  gearing  of  a  lathe  reduces  its  speed  8  times.  The 
cone  pulley  has  5  changes  of  speed.  The  largest  diameter  of 
cone  pulley  on  the  spindle  is  10>£  inches.  The  cone  pulley  on 
the  counter-shaft  is  to  be  of  the  same  size  as  the  cone  pulley  on 
the  spindle,  and  an  even  ratio  of  speed  is  to  be  maintained 
throughout  the  whole  range  of  the  ten  changes  of  speed. 
The  slowest  speed,  when  back  gears  are  in,  is  6  revolutions  per. 
minute.  Calculate  the  speed  of  the  counter-shaft,  the  speed  of 
the  spindle  for  each  change,  and  the  diameter  of  each  step  on 
the  cone  pulley  of  the  spindle. 

Solution : 

The  ratio  of  speed  for  each  step  will  be :. 

.  /7T         — Ql =  — ! =:  0.18062 

V  8  5  5 

The  corresponding  number  is  1.516. 

With  back  gears  in,  the  speed  of  spindle  : 
On  first  cone  is  6  revolutions  per  minute. 
On  second  cone  is  6  X  1.516  =    9  revolutions  per  minute. 
On  third    cone  is  9  X  1.516  =  14  revolutions  per  minute. 
On  fourth  cone  is  14  X  1.516  =  21  revolutions  per  minute. 
On  fifth  cone  is  21  X  1.516  =  32  revolutions  per  minute. 
With  back  gears  out,  speed  of  spindle : 
On  first  cone  will  be    6X8=    48  revolutions  per  minute. 
On  second  cone  will  be    9X8=    72  revolutions  per  minute. 
On  third  cone  will  be  14  X  8  =  112  revolutions  per  minute. 
On  fourth  cone  will  be  21  X  8  =  168  revolutions  per  minute. 
On  fifth  cone  will  be  32  X  8  =  256  revolutions  per  minute. 


PULLEYS. 

The  speed  of  the  counter-shaft  will  be: 

N=  V  48  X  256  =  112  revolutions  per  minute. 

As  the  speed  of  main  lines  in  factories  usually  runs  at  some 
multiple  of  10,  we  may,  for  convenience  in  getting  even-sized 
pulleys  for  connections  between  counter  and  main  shaft,  in 
practical  work,  decide  to  run  the  counter-shaft  110  revolutions 
per  minute. 

(When  a  pair  of  cone  pulleys  has  an  uneven  number  of 
steps,  and  are  cast  from  the  same  pattern,  the  speed  of  the 
counter  should  be  equal  to  the  speed  of  the  machine  when  the 
belt  is  run  on  the  middle  step). 

The  diameter  of  the  largest  step  of  the  cone  pulley  on  the 
spindle  is  10)4  inches.  The  corresponding  step  on  the  counter 

1054  X  48 
will  be  ~iJ(j  —  =  4.581";  practically,  4%"  diameter. 

The  largest  and  smallest  step  on  the  counter-shaft  will  also 
be  10^  and  4^  inches  in  diameter. 

Any  of  the  intermediate  steps  on  the  spindle  may  be  cal- 
culated by  the  formula  : 

-  (P  +  <*)  x  N 


=  9.065  .  practically, 


=  5.932  ;  practically,  6  in. 


Thus,  assuming  the  counter-shaft  to  run  110  revolutions  per 
minute,  the  speed  of  the  spindle,  with  back  gears  out,  on  the  five 
different  steps  will  be  : 


X  10^  —  265  revolutions  per  minute. 

4/£ 

110  X  9 

=  165  revolutions  per  minute. 


6 
110  X 


_  110  revoiutions  per  minute. 


110  X  6     =    73  revolutions  per  minute. 


*    ^   =    47  revolutions  per  minute. 


356  PULLEYS. 

When  the  back  gears  are  in  action  the  speed  will  be : 

Q/»K 

fr  =  33  >£  revolutions  per  minute.  • 

8 

—  =  20  >6  revolutions  per  minute. 

_J_  =  13%  revolutions  per  minute. 
8 

»rq 

_!_  =  9^6  revolutions  per  minute. 
8 

J±Z  =  5  #5  revolutions  per  minute. 
8 

These  speeds  are  all  within  the  practical  requirements  of 
the  problem,  and  now  the  next  operation  is  to  modify  the  diam- 
eters slightly  in  order  to  get  proper  tension  on  the  belt.  (See 
page  352.) 


FLY-WHEELS. 

Fly-wheels  are  used  to  regulate  the  motion  in  machinery  by 
storing  up  energy  during  increasing  velocity,  and  giving  out 
energy  during  decreasing  velocity.  Fly-wheels  cannot  perform 
either  of  these  functions  without  a  corresponding  change  in 
velocity.  The  rim  of  the  wheel  may  be  very  heavy  and  moving 
at  a  high  velocity,  the  change  in  speed  may  be  small  and  hardly 
perceptible  if  the  energy  absorbed  and  given  out  is  small,  but 
there  must  always  be  a  change  in  velocity  to  enable  a  fly-wheel 
to  act.  The  common  expression  of  gaining  power  by  a  heavy 
fly-wheel  is  very  misleading,  to  say  the  least.  There  is  no 
power  gained  by  a  fly-wheel  but,  on  the  contrary,  considerable 
power  is  absorbed  by  friction  in  the  bearings  when  a  shaft  is 
loaded  with  a  heavy  fly-wheel,  (see  example  in  calculating  fric- 
tion, page  305).  Nevertheless,  a  fly-wheel  performs  a  very  use- 
ful function  in  machinery  by  storing  up  energy  when  the  supply 
exceeds  the  demand  and  giving  it  out  at  the  time  it  is  needed  to 
do  the  work.  ( For  momentum  of  fly-wheels  see  example,  page 
300.  For  kinetic  energy,  see  example,  page  301 ). 

Weight  of  Rim  of  a  FIy=Wheel. 

The  weight  of  a  rim  of  a  cast-iron  fly-wheel  will  be : 
W—d^y.  0.7854  X  D  X  3.1416  X  0.26  ;  this  reduces  to, 
W=D  X  d2  X  0.64 
D  =  Middle  diameter  of  rim  in  inches. 
d—  Diameter  of  section  of  rim  in  inches. 
W  =  Weight  in  pounds. 


FLY-WHEELS.  35  7 

EXAMPLE. 

A  round  rim  of  a  fly-wheel  is  4  inches  in  diameter  and  the 
middle  diameter  of  the  -wheel  is  36  inches.  What  is  the  weight 
of  the  rim  ? 

Solution: 

^=36X4X4X  0.64  =  369  pounds. 

For  a  rim  of  rectangular  section  the  weight  will  be  : 
W  —  Width  X  thickness  X  D  X  3.1416  X  0.26 
IV  =  Width  X  thickness  X  D  X  0.816 

EXAMPLE. 

The  width  of  the  rim  is  six  inches,  the  thickness  is  two 
inches,  and  the  middle  diameter  of  the  rim  is  48  inches.  What 
is  the  weight  of  the  rim  ? 

Solution  : 

W  —  2  X  6  X  48  X  0.816  =  470  pounds. 

Centrifugal  Force  in   Fly-Wheels  and  Pulleys. 

Pulleys  are  not  only  liable  to  be  broken  by  the  stress  due  to 
the  action  of  the  driving  belt,  but  in  fast-running  pulleys  and 
fly-wheels  the  stress  due  to  centrifugal  force  is  far  more  dan- 
gerous. This  stress  increases  as  the  square  of  the  velocity  and 
directly  as  the  weight,  therefore  there  is  a  limit  to  the  velocity 
at  which  fly-wheels  and  pulleys  can  be  run  with  safety. 

Generally  speaking,  increasing  the  thickness  of  the  rim 
does  not  increase  its  strength,  because  the  total  tensile  strength, 
the  total  weight  of  the  rim,  and,  consequently,  also  the 
centrifugal  force,  increase  in  the  same  proportion;  but  it  has 
great  influence  upon  the  strength  of  the  wheel  to  have  the  ma- 
terial in  the  rim  distributed  to  the  best  advantage.  At  the  same 
time  it  is  very  important  to  construct  the  rim  and  arms  of  such 
proportions  that  the  initial  stress  due  to  uneven  cooling  in  the 
foundry,  is  avoided. 

The  common  formula  is  : 

Mass  X   (velocity)2 
Centrifugal  force  =   -         ----  -v.  ---- 

n  X  r  X  2  K 


Ma^        Weight 

w  (n  x 

Velocit 
r  X  2  TT  V 

32.2 
Therefore, 

ct  - 

-       V 

60           / 

60 


32.2  X  r 
W  X  «2  X  r2  X  0.01096628 
^  '  32.2  X  r 

<:/  =  W  X  «2  X  r  X  0.00034 
^/  •=  Centrifugal  force  in  pounds. 


358 


FLY-WHEELS. 


W  =  Weight  of  revolving  body  in  pounds. 
n  =  Number  of  revolutions  per  minute. 
r  —  Middle  radius  of  pulley  rim  in  feet. 
Thus,  for  any  body  'whose  center  of  gravity  swings  in  a 
circle  of  one  foot  radius,  at  a  speed  of  one  revolution  per  min- 
ute, the  centrifugal  force  will  be  0.00034  times  the  weight  of 
the  body. 

EXAMPLE. 

The  rim  of  a  fly-wheel  is  five  feet  in  middle  radius  and 
weighs  8000  pounds.  It  makes  75  revolutions  per  minute. 
What  is  the  stress  due  to  centrifugal  force? 

Solution : 

Centrifugal  force  =  8000  X  752  X  5  X  0.00034  =  76500  pounds. 

PIG  3>  This  is   the   total  centrifugal  force 

~ — i — *w  tending  to  burst  the  rim,  (see  arrows  in 

/\    A    /\  Fig.  3);  the  force   tending  to >  tear  the 

Xs^  ^A        rim  asunder  in  any  two  opposite  points 

a._L*  —1-6     as  a,  b,  is  3  ^Q  x  2  =  12175  pounds. 

\+  ^J  The  next  question  is :     Has  the  sec- 

X    *          \7        tion  of  the  rim  tensile  strength  enough  to 

^<JL,^  resist  this  stress  with  safety?     If  not, 

either  decrease  the  rim  speed  or  make  the  rim  of  material  having 

more  tensile  strength. 

The  centrifugal  force  for  the  same  number  of  revolutions 
increases  as  the  radius,  therefore  the  average  centrifugal  force 
acting  in  the  arms  is  only  about  half  of  the  centrifugal  force 
acting  in  the  rim,  and  as  the  stretch  is  in  proportion  to  the  stress, 
the  rim  tends  to  stretch  more  than  the  arms,  and,  consequently, 
it  can  not  yield  freely  to  the  action  of  the  centrifugal  force, 
but  is  to  a  certain  extent  held  back  at  the  junction  with  the 
arms.  This  action  is  shown  in  an  exaggerated  form  at  a,  Fig.  4. 
In  regard  to  this  action,  the  part  of  the  rim  between  the 

arms  may  be  considered 
as  a  beam  fastened  at  both 
ends  and  uniformly  loaded 
throughout  its  whole  length 
equal  to  that  amount  of  cen- 
trifugal force  in  the  rim 
which  is  resisted  by  the 
arms ;  therefore,  the  rims  of 
large  pulleys  should  always 
be  ribbed  on  the  inside. 
(See  cross-section  of  rim  at 
X,  Fig.  4). 

Another  bad  feature  frequently  seen  in  pulleys  is  the 
counter-balance.  (See  b,  Fig.  4).  This  little  piece  itself,  weigh- 
ing probably  only  five  pounds,  holds  the  pulley  neatly  in  balance 


FIG  4 


_ 


PLY- WHEELS.  359 

and  is  very  innocent  as  long  as  the  pulley  is  standing  still,  but 
imagine  what  stress  it  will  produce  on  the  rim  of  a  6-foot  pulley 
running  at  a  rim  speed  of  80  feet  per  second. 
Solution : 

c  f= =  331  pounds. 

3  X  32.2 

Thus,  when  that  pulley  is  running  at  a  speed  of  80  feet  per 
second  this  counter-balance  of  five  pounds  will  produce  the 
same  stress  as  if  it  was  loaded  with  331  pounds  when  standing 
still;  therefore,  it  is  evident  how  important  it  is  to  turn  fast- 
running  pulleys  both  inside  and  outside  in  order  to  reduce 
counter-balancing  to  the  least  possible  amount. 

The  danger  of  the  rim  deflecting  or  breaking  from  the  stress 
due  to  the  resistance  from  the  arms  (as  shown  in  Fig.  4),  can  be 
avoided  by  running  ribs  on  the  inside  of  the  rim,  and  the  danger 
caused  by  counter-balancing  can  be  entirely  eliminated  by  mak- 
ing the  pulley  balance  without  adding  any  balancing  pieces. 
Thus,  one  danger  of  breaking  is  avoided  by  proper  designing 
and  the  other  by  good  workmanship. 

The  direct  action  of  the  centrifugal  force  on  the  rim  is  cal- 
culated by  the  formula, 

cf—  iu  X  n2  X  r  X  0.0034, 

and  the  weight  of  the  rim  of  a  cast-iron  fly-wheel  having  one 
square  inch  of  sectional  area  and  a  radius  of  one  foot  will  be 
2  X  TT  X  12  X  0.26  pounds,  and  a  ring  of  r-foot  radius  will 
weigh  ;-  X  2  X  TT  X  12  X  0.26  pounds.  As  already  stated,  the 
centrifugal  force  increases  as  the  square  of  the  velocity ;  that 
is,  if  the  number  of  revolutions  is  doubled  the  centrifugal  force 
is  increased  four  times ;  thus  is  the  dangerous  limit  approached 
very  rapidly  under  increased  speed,  and  in  order  to  prevent 
accident,  if  the  speed  should  happen  to  increase,  it  is  necessary 
always  to  use  a  high  factor  of  safety  in  such  calculations. 
Thus,  using  15  as  a  factor  of  safety*  and  assuming  the  tensile 
strength  of  cast-iron  as  12,000t  pounds  per  square  inch,  the 
stress  in  each  cross-section  at  a  and  b  must  not  exceed  800 
pounds  per  square  inch.  The  allowable  centrifugal  force  in 
both  sections  of  the  rim  is  2  X  800  pounds,  and  inserting  those 
values  and  solving  for  n  the  greatest  number  of  revolutions 
allowable  for  a  cast-iron  fly-wheel  will  be : 

2  TT  X  12  X  0.26  X  r  X  n2  X  r  X  0.000340568  =  800  X  2 
n*  r*  =  752891 

n  r  =  \/752891 
n  =  868  r  =  868 

r  n 


*  15  as  factor  of  safety  with  regard  to  strength,  is  only  yl5  =  3.873,  or 
ICES  than  4,  as  factor  of  safety  with  regard  to  speed. 

t  This  tensile  strength  for  cast-iron  may  seem  very  low,  but  it  is  dangerous  to 
assume  more,  because  ot  inside  stress  in  arms  or  rim  already,  due  to  uneven  cool- 
ing of  the  casting  in  the  foundry. 


360  FLY-WHEELS. 

Transposing  this  to  diameter  in  inches,  the  constant  will  be 
24  X  868  —  20832.     The  formula  will  be : 

Number  of  revolutions  per  minute  = 


Diameter  in  inches. 


Diameter  in  inches  =  20832 


Revolutions  per  minute. 

Rule  for  Calculating  Safe  Speed. 

Divide  20832  by  the  diameter  of  the  fly-wheel  in  inches,  and 
the  quotient  is  the  allowable  number  of  revolutions  per  minute 
at  which  a  well-constructed  fly-wheel  may  be  run  with  safety. 

Rule  for  Calculating  Safe  Diameter. 

Divide  20832  by  the  number  of  revolutions  per  minute,  and 
the  quotient  is  the  safe  diameter  in  inches  for  a  well-constructed 
fly-wheel. 


SHAFTING. 

When  calculating  strength  of  shafting,  both  transverse  and 
torsional  stress  should  be  considered.  Transverse  stress  is 
produced  by  the  weight  of  the  shaft  itself,  the  pulleys  and  the 
tension  of  the  belts,  the  effect  of  which  is  very  severe  if  the 
distance  between  the  hangers  is  too  long.  Torsional  stress  is 
produced  by  the  power  which  the  shaft  transmits.  Usually  the 
distance  between  the  hangers  is  made  so  short  that  the  torsional 
stress  on  a  shaft  is  the  greater.  For  transverse  stress  the  shaft 
may  be  considered  as  a  round  beam,  supported  under  the  ends  and 
loaded  somewhere  between  supports.  According  to  Table  No. 
30  the  transverse  stress  which  will  destroy  a  wrought  iron  beam 
one  inch  square,  fastened  at  one  end  and  loaded  at  the  other,  is 
600  pounds ;  the  strength  of  a  round  beam  of  the  same  diameter 
is  (see  page  251)  0.6  that  of  a  square  beam.  When  the  beam 
is  supported  under  both  ends  and  loaded  in  the  middle,  its 
breaking  load  will  increase  four  times ;  therefore,  the  constant, 
c,  will  be  600  X  4  X0.6  —  1440.  Using  10  as  factor  of  safety, 
the  formula  for  transverse  strength  of  a  wrought  iron  shaft  will 
be: 

L  _  144  Z>3  Trr_   144  Z>3 


w 

D  =  Diameter  of  shaft  in  inches. 
L  =  Distance  between  hangers  in  feet. 
W  =  Transverse  load  in  pounds,  supposed  to  be  at  the 

middle,  between  the  hangers. 

144  =  Constant  for  wrought  iron,  and  100  to  120  may  be 
used  as  constant  for  cast-iron,  with  10  as  factor  of  safety  for 
transverse  strength^ 


SHAFTING.  361 

Formula  1,  expressed  as  a  rule,  will  be: 

Multiply  the  distance  between  hangers,  measured  in  feet, 
by  the  transverse  load  in  pounds  ;  divide  this  product  by  144,  and 
the  cube  root  of  the  quotient  will  be  the  diameter  of  the  shaft 
in  inches,  calculated  with  10  as  factor  of  safety  for  transverse 
strength. 

Shaft  not  Loaded  at  the  Middle  Between  the  Hangers. 

When  a  shaft  is  not  loaded  at  the  middle  of  the  span,  but 
somewhere  toward  one  of  the  hangers,  it  will  carry  a  heavier 
load,  with  the  same  degree  of  safety,  than  it  would  if  loaded  in 
the  middle,  and  the  ratio  is  in  inverse  proportion  as  the  square 
of  half  the  distance  between  hangers  to  the  product  of  the  short 
and  the  long  ends  of  the  shaft.  For  instance,  a  shaft  is  six  feet 
between  hangers  and  loaded  at  the  middle.  What  would  be  the 
difference  in  transverse  strength  if  it  was  loaded  two  feet  from 
one  hanger  and  four  feet  from  the  other? 

3X3  =  9  and  2X4  =  8. 

Thus,  find  the  transverse  load  for  a  shaft  when  loaded  in 
the  middle,  multiply  by  9  and  divide  by  8,  and  the  quotient  is 
the  load  which  the  same  shaft  will  carry  with  the  same  degree 
of  safety  against  transverse  stress,  if  loaded  two  feet  from  one 
end  and  four  feet  from  the  other. 

This  rule  only  applies  to  the  transverse  strength,  and  not  to 
the  transverse  stiffness  of  the  shaft.  For  different  shapes  of 
shafts  and  different  modes  of  loading,  see  beams,  pages  243-244. 
When  shafts  are  heavily  loaded  near  one  hanger,  and  the  hanger 
on  the  other  side  of  the  pulley  is  further  off,  most  of  the  load  is 
thrown  on  the  bearing  nearest  to  the  pulley,  and  this  bearing  is, 
therefore,  liable  to  heat  and  to  cause  trouble,  even  if  the  shaft 
is  both  stiff  and  strong  enough.  (  See  reaction  on  the  support 
of  beams,  page  252). 

Transverse  Deflection  in  Shafts. 

The  transverse  deflection  in  a  shaft  may  be  calculated  by 
the  formula: 


L  = 

L*WC 


~L*~C 


S  —  Deflection  in  inches. 
D  =  Diameter  of  shaft  in  inches. 

L  =  Length  of  span  in  feet. 
W  =  Load  on  middle  of  shaft  in  pounds. 

C—  Constant  =  1.7  X  constant  in  Table  No.  31,  and  for 
wrought  iron  or  Bessemer  steel  may  be  taken  as  0.00002652. 


362  SHAFTING. 

Constant  C  may  be  calculated  from  experiments  by  the 
formula, 

C=T^W 
S  =  Deflection  in  inches  noted  in  the  specimen,  when 

supported  under  both  ends  and  loaded  transversely 

at  the  middle  between  supports. 
D  =  Diameter  of  specimen  in  inches. 
L  =  Distance  between  supports  of  specimen  in  feet. 
W=  Experimental  load  in  pounds. 

EXAMPLE. 

A  round  specimen  placed  in  a  testing  machine,  supported 
under  both  ends  and  loaded  at  the  middle  with  2000  pounds, 
deflects  0.1  inch.  The  diameter  of  the  specimen  is  two  inches 
and  the  distance  between  supports  is  three  feet.  Calculate 
constant  C  for  this  kind  of  material. 
Solution : 

c  _    0.1  X  2* 
33  X  2000 

C  =      L6     =  0.0000296  inch. 
54000 

Thus,  the  deflection  for  this  kind  of  material  is  0.0000296 
inch  per  pound  of  load,  applied  at  the  middle,  between  supports, 
for  a  round  bar  one  inch  in  diameter  and  one  foot  between 
supports. 

Allowable  Deflection  in  Shafts. 

The  distance  between  the  hangers  must  always  be  deter- 
mined with  due  consideration  to  the  allowable  transverse  deflec- 
tion in  the  shafting,  especially  when  the  shaft  is  loaded  with 
large  pulleys  and  heavy  belts,  remembering  that  the  deflection 
increases  directly  with  the  transverse  load  and  with  the  cube  of 
the  length  between  the  bearings,  (see  page  254).  The  allowable 
transverse  deflection  in  shafting  ought  not  to  exceed  0.006  to 
0.008  inch  per  foot  of  span  ( see  page  266).  A  beam  of  wrought 
iron  one  foot  long  and  one  inch  square,  when  supported  under 
both  ends  and  loaded  at  the  middle,  will  deflect  0.0000156  inch 
per  pound  of  load,  (see  Table  No.  31,  page  259),and  a  round  beam 
deflects  1.7  times  as  much  as  a  square  beam,  when  the  diameter 
and  side  are  equal.  A  round  shaft,  one  inch  in  diameter  and 
one  foot  long,  when  loaded  at  the  middle  with  144  pounds  will, 
therefore,  deflect  144  X  1.7  X  0.0000156  =  0.00382  inch. 

Thus,  this  load  does  not  give  more  than  an  allowable  de- 
flection. But,  suppose  the  distance  between  bearings  is  doubled 
and  the  load  decreased  one-half;  the  ultimate  strength  of  the 
shaft  will  be  the  same,  but  the  deflection  will  be  72  X  1.7  X  23  X 
0.0000156  =  0,01528  =  0.0764  inch  per  foot. 


SHAFTING.  363 

This  calculation  shows  plainly  how  very  necessary  it  is  to 
have  bearings  near  the  pulleys  where  shafts  are  loaded  with 
heavy  pulleys  and  large  belts.  There  is  nothing  more  liable  to 
destroy  a  shaft  than  too  much  deflection,  because  the  shaft  is, 
when  running,  continually  bent  back  and  forth,  and  at  last  it 
must  break.  The  fact  must  never  be  lost  sight  of  that  strength 
and  stiffness  are  two  entirely  different  things  and  follow  entirely 
different  laws;  therefore,  after  calculations  are  made  for 
strength,  the  stiffness  must  also  be  investigated,  as  stiffness  is 
a  very  important  property  in  shafting.  The  best  way  to  over- 
come too  much  transverse  deflection  is  to  shorten  the  distance 
between  the  bearings.  Of  course,  increasing  the  diameter  of  the 
shaft  will  also  overcome  deflection,  but  shafting  should  never  be 
larger  in  diameter  than  necessary,  because  the  first  cost  increases 
with  the  \yeight,  which  increases  as  the  square  of  the  diameter, 
and  the  frictional  resistance  will  also  increase  with  the  increased 
diameter  ;  consequently,  also,  the  running  expenses. 

Torsional  Strength  of  Shafting. 

Shafting  may  be  considered  as  a  beam  fastened  at  one  end 
and  having  a  torsional  load  applied  at  the  other  end  equal  to 
the  pull  of  the  belt  on  an  arm  of  the  same  length  as  the  radius  of 
the  pulley.  In  Table  No.  32,  page  268,  constant  c  is  given  as  580 
pounds  for  wrought  iron. 

The  formula  for  twisting  stress,  as  explained  under  beams 
(  see  page  267)  is, 

P  =    D*c  D  =  JZ*I 

m  \      ~c 

;;/  =  the  length  of  the  lever  or  arm  in  feet,  and  will  here 
be  the  radius  of  the  pulley  and  be  denoted  by  r.  The  length  of 
the  shaft  has  no  influence  on  its  torsional  strength,  but  only  on 
its  angle  of  torsional  deflection  (see  page  268).  Using  10  as 
factor  of  safety,  the  formula  will  be  : 


58 

D  =  Diameter  of  shaft  in  inches. 
r  =  Radius  of  pulley  in  feet. 

W=  Pull  of  belt  in  pounds. 

58  =  Constant,  with  10  as  factor  of  safety  =  Vio  X  580, 
taken  from  Table  No.  32,  page  268. 

Frequently  it  is  more  convenient  to  calculate  the  torsional 
strength  of  shafting  according  to  the  number  of  horse-power  the 
shaft  is  to  transmit  (  see  page  317  ). 

In  the  above  formula,  assume  W  to  be  58  pounds,  r  to  be 
one  foot,  and  D  will  be  one  inch.  That  is,  a  shaft  one  inch  in 
diameter  is  strong  enough  to  resist,  with  10  as  factor  of  safety, 


364  SHAFTING. 

a  torsional  load  of  58  pounds  acting  on  an  arm  one  foot  long. 
Assuming  this  58  pounds  to  act  on  the  rim  of  a  pulley  of  one 
foot  radius,  two  feet  in  diameter,  and  making  one  revolution  per 
minute,  it  will  transmit  power  at  a  rate  of  58  X  6%  =  364  f  foot- 
pounds per  minute  ;  but  one  horse-power  is  33,000  foot-pounds 
per  minute,  and  if  the  shaft  should  transmit  one  horse-power  it 

33000 
must  make    0^44     —  90.52  revolutions  per  minute.     Hence  the 

practical  formulas  for  torsional  strength  of  shafting  : 


90 


H  X  90 


90 


D  —  Diameter  of  shaft  in  inches. 

H  =•  Number  of  horse-power  transmitted  by  the  shaft. 
11  •=.  Number  of  revolutions  made  by  the  shaft  per  minute. 
90  —  Constant,  using  10  as  factor  of  safety,  and  assuming 
the  torsional  strength  to  be  as  given  in  Table  No.  32. 


Torsional  Deflection  in  Shafting. 

In  constructing  different  kinds  of  machinery  it  is  frequently 
necessary  to  consider  the  torsional  deflection.  The  formula 
for  torsional  deflection  for  wrought  iron  ( see  page  271)  will  be  : 

c  _  0.00914  X  m  L  P 

— £>4 —  This  will  transpose  to 

s_  48  H  L 

'~n D± 

S  =  Deflection  in  degrees. 
H '  —  Number  of  horse-power  transmitted. 

0.00914  X  33000  _ 
48  —  Constant ;  calculated  thus,  — 2X3  1416 — 

L  =•  Length  of  shaft  in  feet  between  the  force  and  the 

resistance. 

n  =  Number  of  revolutions  made  by  the  shaft  per  minute. 
D  =  Diameter  of  shaft  in  inches. 
EXAMPLE. 

How  many  degrees  is  the  deflection  of  a  shaft  two  inches 
in  diameter,  50  feet  long,  making  300  revolutions  per  minute  and 
transmitting  15  horse-power,  applied  at  one  end  and  taken  off  at 
the  other? 
Solution : 

48  HL  _  48  X  15  X  50   _ 
^  -  ~^~D±~        ~  300  X  2*         ~  7/2 


SHAFTING. 


365 


Classification  of  Shafting. 

Shafting  may  be  divided  into  three  different  kinds. 

First. —  Shafts  where  the  main  belts  are  transmitting  the 
power,  or  so-called  "  Jack  Shafts."  Such  shafts  must  have  their 
boxes  as  near  the  pulleys  as  possible.  For  torsional  strength 
their  diameter  may  be  calculated  by  the  formula, 


D  =  V— *—          (See  Table  No.  40.) 

Second. —  Common  shafting  in  shops  and  factories,  where 
the  power  is  taken  off  at  different  places  for  driving  machinery. 
Such  shafts  ought  to  be  supported  by  hangers  as  given  in 
Table  No.  43,  and  their  supports  must  also  be  reinforced  by 
extra  hangers,  if  necessary,  where  an  extraordinary  large  pulley 
or  heavy  belt  is  carried.  For  torsional  strength  the  diameter  of 
such  shafts  may  be  calculated  by  the  formula, 


=  \--^-  (See  Table  No.  41.) 

Third. —  Shafting  having  practically  no  transverse  stress, 
but  used  simply  to  transmit  power  from  one  place  to  another. 
Such  shafts  ought  to  be  supported  by  hangers  according  to 
Table  No.  43,  and  the  diameter  may  be  calculated  by  the 
formula, 


-  ^  \  H  X  50 


(See  Table  No.  42.) 


TABLE  No.   40.— Giving  Horse-Power  of  Main  Shafting 
at  Various  Speeds. 


III 

Revolutions  per  Minute. 

!-°i 

60 

80 

100 

125 

150 

175 

200 

225 

250 

275 

300 

1V 

2.6 

3.4 

4.3 

5.4 

6.4 

7.5 

8.6 

9.7 

10.7 

11.8 

12.9 

•2 

3.8 

5.1 

6.4 

8 

9.6 

11.2 

12.8 

14.4 

16 

17.6 

19 

5.4 

7.3 

9.1 

11 

13 

16 

18 

21 

23 

25 

27 

2  y 

7.5 

10 

12.5 

15 

18 

22'  25 

28 

31 

34 

37 

2% 

10 

13 

16 

21 

25 

29 

33 

37 

42 

46 

50 

3 

13 

17 

21 

27 

32 

38 

43 

49 

54 

59 

65 

16 

22 

27 

34 

41 

48 

55 

62 

69 

76 

82 

35^ 

20 

•21 

34 

43 

51 

60 

68 

77 

86 

94 

103 

337 

25 

:!4 

42 

53 

63 

74 

84 

95 

105 

116 

126 

4 

30 

41 

51 

64 

77 

IK)  102 

115 

128 

141 

154 

5  2 

43 

60 

58 
80 

73 
100 

91 
125 

109 
150 

128 
175 

146  164 

200  1225 

182 
250 

201 
275 

219 
300 

366 


SHAFTING. 


TABLE  No.  41.— Giving  Horse=Power  of  Line  Shafting 
at  Various  Speeds. 


Ill 

Revolutions  per  Minute. 

§11 
SO.H 

100 

125 

150 

175 

200 

225 

250 

275 

300 

325 

350 

400 

1* 

6 

7.4 

9 

10.4 

12 

13.4 

15 

16.4 

18 

19.4 

21 

23.8 

i# 

7.3 

9.1 

10.9 

12.7 

14.5 

16 

18 

20 

22 

23.8 

25 

29 

2 

8.9 

11.1 

13 

15.5 

17.7 

20 

22 

24 

27 

19 

31 

35 

2/1? 

10.6 

13.2 

16 

18.5 

21 

24 

27 

29 

32 

34 

37 

42 

2X 

12.6 

15.8 

19 

22 

25 

28 

32 

35 

38 

41 

44 

50 

2^ 

15 

18.6 

22 

26 

30 

33 

37 

41 

44 

48 

52 

59 

2^ 

17 

22 

26 

30 

35 

39 

43 

48 

52 

56 

61 

69 

2% 

23 

29 

34 

40 

46 

52 

58 

64 

69 

75 

81 

92 

3 

30 

37 

45 

52 

60 

67 

75 

82 

90 

97 

105 

120 

3X 

38 

47 

57 

67 

76 

86 

95 

105 

114 

124 

133 

152 

3/^ 

48 

59 

71 

83 

95 

107 

119 

131 

143  1  155 

L67 

190 

3|^ 

58 

73 

88 

102 

117 

132 

146 

161 

L76  190 

205 

234 

4 

71 

89 

107 

125 

142 

160 

178 

196 

213  231 

249 

284 

TABLE  No.  42.—  Giving  Horse=Power  of  Shafting  Used 

Only  for  Transmitting  Power. 

isi 

Revolutions  per  Minute. 

fil 

100 

125 

150 

175 

200 

225 

250 

275 

300 

32E 

35( 

400 

ll/2 

6.7 

8.4 

10 

11.8 

13.5 

15.1    16.8 

18.E 

20.3 

22 

2c 

27 

1.^$ 

8.6 

10.7 

12.8 

15 

17.1 

19.3    21.4 

23.6 

25.8 

28 

3C 

34 

1% 

10.7 

13.4 

16 

18.7 

21.5 

24 

26.8 

29.4 

32.1 

35 

3-5 

43 

i^ 

13.2 

16.5 

19.7 

23 

26.4 

29.7    33 

36.2 

39.5 

43 

4fc 

52 

2 

16 

20 

24 

28 

32 

36       40 

44 

48 

52 

56 

64 

2/^ 

19 

24 

29 

33 

38 

42    j  48 

53 

57 

62 

67 

76 

2X 

23 

28 

34 

40 

45 

51 

57 

63 

68 

74 

80 

91 

2^ 

27 

33 

40 

47 

54 

60 

67 

74 

80 

87 

94 

107 

2^ 

31 

39 

47 

55 

62 

70 

78 

86 

94 

102 

109 

125 

2|^ 

41 

52 

62 

73 

83       93 

104 

114 

125 

132 

146 

166 

3 

54 

67 

81 

94 

108 

121 

135 

148 

162 

175 

189 

216 

3% 

69 

86 

103 

120 

137 

154 

172 

189 

206 

223 

240 

275 

l# 

86 

107 

128 

150 

171 

193 

114 

236 

257     279 

300 

343 

SHAFTING. 


367 


Distance  Between  the  Bearings. 

Jack  shafts  should  always  have  bearings  as  near  the  pulleys 
as  possible. 

Ordinary  line  shafts,  as  given  in  Table  No.  41,  and  shafts 
for  simply  transmitting  power,  may  have  the  distance  between 
the  hangers  as  given  in  the  following  table  : 

TABLE  No.  43- 


Diameter  of  Shaft  in  Inches. 

IX  tol# 

2  to2^ 

2^  to  4 

Distance  between  bearings  in  feet 

6/2 

8 

10 

to  be  transmitted  from  A  to  B 
The  gears  on  shafts  A  and  B 

FIG.  1. 


Shafts  for  Idlers. 

Shafts  for  idlers  ( see  C,  Fig.  1 )  have  very  little  torsional 
stress  and  the  distance  between  the  bearings  may  also  be  very 
short,  so  that  even  with  a  great  transverse  load  such  a  shaft 
may  be  of  comparatively  small  diameter  as  far  as  requirements 
for  strength  is  concerned.  In  such  a  shaft  there  is  great  danger 
of  trouble  from  hot  bearings ;  therefore,  in  designing,  it  is  very 
important  to  make  its  diameter  and  the  length  of  the  bearing  of 
such  proportions  that  excessive  pressure  per  square  inch  of 
bearing  surface  is  avoided. 

EXAMPLE. 

Twenty-five  horse-power  is 
through  idler  C.  (See  Fig.  1). 
are  36  inches  in  diameter  and 
make  40  revolutions  per  minute. 
What  is  the  necessary  diameter 
of  shaft  C,  which  is  supported 
by  two  bearings  one  foot  apart 
and  carrying  a  gear  48  inches 
in  diameter  placed  at  the  mid- 
dle between  the  bearings. 

Solution : 

The  velocity  on  pitch  line  of  gear  A  will  be 

40  X  36  X  3.1416  =  ^  f  minute 

12 

25  horse-power  =  33,000  X  25  —  825,000  foot-pounds. 
The  pressure  at  the  pitch  line  of  A  transferred  to  Cwill  be 

825000  =  2188  pounds. 
377 


SHAFTING. 

The  reaction  at  the  pitch  line  between  C  and  B,  is  also 
2188  pounds;  therefore,  the  total  pressure  (besides  the  weight 
of  C,  which  is  omitted  in  this  calculation  )  on  both  bearings  will 
be  2  X  2188  =  4376  pounds  and  the  pressure  of  each  bearing  of 
C  will  be  2188  pounds.  Allowing  a  pressure  on  the  bearings  of 
100  pounds  per  square  inch,  the  necessary  bearing  surface 
will  be 

218S 

-  =  21.88  square  inches  for  each  bearing. 

J.UU 

Assuming  the  length  of  the  bearing  to  be  twice  its  diameter, 
D  X  2  D  —  21.88 
ja-21.88 


Calculating  the  size   required    with  regard  to   transverse 
strength  by  the  formula  on  page  360, 

3 


=  J 

\ 


-J1X487U  =8.12  inches. 
144 


Thus,  a  shaft  3.3  inches  in   diameter  is  of  ample  size  for 
strength.     The  surface  velocity  of  this  shaft  will  be, 

8.3  X  3.1416  X  40  =  34.feet 

per  minute,  and  at  that  velocity  a  pressure  of  100  pounds  per 
square  inch  of  bearing  surface  is  very  safe  from  liability  of 
heating  if  the  bearing  is  well  made  and  amply  provided  with  oil. 


Proportion  of  Keys. 

The  breadth  of  the  key  is  usually  made  to  be  one-fourth  of 
the  diameter  of  the  shaft,  and  the  thickness  to  be  one-sixth  of 
the  diameter  of  the  shaft. 

Keys  and  key-ways  are  usually  made  straight  and  should 
always  be  a  very  good  fit  sidewise.  Frequently  set-screws  are 
used  on  top  of  keys  in  mill  gearing.  Sometimes  in  heavy  ma- 
chinery keys  are  made  tapering  in  thickness,  usually  one-eighth 
inch  per  foot  of  length.  A  corresponding  taper  is  made  in  the 
depth  of  the  key-way  in  the  hub.  Key-ways  in  shafts  are  always 
made  straight. 

For  light  and  fine  machinery  taper  keys  are  never  used* 


SHAFTING. 


TABLE  No.  44.— Dimensions  of  Couplings  for  Shafts. 

(All  dimensions  in  inches.) 


Diameter 
of  Shaft. 

Dimensions  of  Couplings. 

Diameter  of 

Number  of 

d 

D 

L 

/ 

** 

2  2 
3  2 
4  2 
5  2 

3 

5  2 
6 

7* 
9  * 

7 
8 
9 
10 

13  2 

14/2 

15* 

17 

4 
6  2 

6^: 

7# 

8* 

¥ 

n 
i 

H 
X 

8 

^ 

i  8 

4 
4 
4 
4 
5 
5 
0 
6 
6 

1  1"-  1 

^  ^ 

i 

± 

i 

m^ 

j 

^  —  . 

BE^ 

-  d—  H 
VRINQS. 

A  satisfactory  rule  is  to  make  the  length  of  the  bearing  for 
line  shafting  six  times  the  square  root  of  the  diameter  of  the 
shaft. 

EXAMPLE. 

What  is  a  suitable  length  of  bearings  for  a  shaft  of  four 
inches  diameter? 

Solution : 

Length  of  bearing  =  C  X  *ST~=  12  inches. 
Some  designers  make  the  length  of  the  bearing  four  times 
its  diameter. 

Area  of  Bearing  Surface. 

The  projected  area  of  any  bearing  is  always  considered  as 
its  bearing  surface.  Thus,  the  length  of  the  bearing  multiplied 
by  the  diameter  of  the  shaft  gives  the  area  of  bearing  surface. 
For  instance,  the  length  of  the  box  is  twelve  inches  and  the 
diameter  of  the  shaft  is  four  inches ;  the  area  of  bearing  surface 
is  12  X  4  =  48  square  inches. 

Allowable  Pressure  in  Bearings. 

The  allowable  pressure  per  square  inch  of  bearing  surface 
will  depehd  on  the  surface  speed  of  the  shaft  and  the  condition 


370 


BEARINGS. 


of  the  bearing,  arrangements  for  oiling,  etc.  For  common  line 
shafting  from  two  to  four  inches  in  diameter,  not  making  over 
200  revolutions  per  minute,  a  pressure  not  exceeding  forty 
pounds  per  square  inch  ought  to  work  well.  Greater  pressure 
or  greater  speed  may  make  it  difficult  to  keep  the  bearings  cool. 

EXAMPLE. 

What  pressure  may  be  allowed  on  a  bearing  twelve  inches 
long  and  four  inches  in  diameter  ? 

Solution : 

Pressure  =  4  X  12  X  40  =  1920  pounds. 

In  well  constructed  machinery  there  should  not  be  any 
trouble  from  heating,  if  the  surface  velocity  and  the  pressure  in 
the  bearings  does  not  exceed  the  values  given  in  the  following 
table  :— 


METRIC  MEASURE. 


ENGLISH  MEASURE. 


Kilograms  per  Square 
Centimeter. 

Surface  Velocity  in 
Meters  per  Minute. 

Pounds  per  Square 
Inch. 

Surface   Velocity   in 
Feet  per  Minute. 

5 
12 

20 

100 
50 
20 

75 

180 
300 

300 
150 
60 

The  bearings  for  machinery  in  general  are  constructed  in 
various  ways  and  of  different  proportions,  according  to  the  de- 
signer's judgment,  but  it  is  a  well-known  fact  that  high-speed 
machinery  must  have  longer  bearings  than  slow-speed  ma- 
chinery. 

The  length  of  the  bearing  will  usually  vary  from  one  and 
one-half  to  six  times  the  diameter. 

When  the  shafts  are  small  (less  than  two  inches  in  diameter), 
and  the  speed  is  from  100  to  1000  revolutions  per  minute,  the 
following  empirical  formula  may  be  used  as  a  guide  : 


L  =  length  of  bearing,  d—  diameter  of  bearing,  n  =  num- 
ber of  revolutions  per  minute. 

FIG.    1. 


BEARINGS. 


37* 


FIGURE  1  shows  a  cheap,  solid  cast-iron  box  used  for  com- 
paratively small  and  less  important  shafts.  Dimensions,  suita- 
l)le  for  bearings  from  one  to  two  inches  in  diameter,  are  given 
in  the  following  table  : — 

(All  Dimensions  in  Inches.) 


55 


73 

bJ3  bJO 
C   C  ^ 

(11    **""        >* 


25/^ 
3A 


2A 


334 

6  4 
6^4 


« 


5/8 
« 


m 


FIGURE  2  shows  a  babbitted  split  box  suitable  for  shafts 
from  one  to  four  inches  in  diameter,  and  running  at  a  com- 
paratively slow  speed. 


FIG.  2. 

d=  Diameter  of  shaft. 

a  —  2)^  X  d. 
b—\#Kd. 

c  —  3  X  d  -f-  %  inch. 

k  —  4  X  </+  1^  inches. 

*  =  X  X  d  +  %  inch. 

/=  #  X  ^/+  Xinch. 

^=1^  X  d. 

/=2  X  ^ 

Thickness  of  babbitt  metal,  t=V\Qd+  %  inch. 
Diameter  of  bolt,  //  •=•  y$d  +   ^  inch. 
Diameter  of  bolt,  /'  =  Vzd  -{-  ^  inch. 

FIGURE  3  shows  the  same  general  design  of  box  as  Figure 
2,  excepting  that  the  bearing  is  longer  and  the  base  wider. 
This  box  is  more  suitable  for  comparatively  high-speed  shafts. 


BEARINGS. 


FIG.  3. 


=  Diameter  of  shaft. 
IX  inches. 

inch. 

c  —  2y2d-\-  2  inches. 
k  —  3  d  +  3  inches. 
<?==  X</  +  X  inch. 


r=5  X  V  </ 


i  —  y%  d  +  ft  inch. 


FIGURE  4  shows  a  babbitted  box  or  pedestal  suitable  for 
comparatively  heavy-loaded  shafts,  from  three  to  eight  inches 
in  diameter,  such  as  outer  bearings  for  steam  engine  shafts, 
bearings  for  jack  shafts,  etc. 


Flo.  4. 


d—  Diameter  of  shaft;  <z  =  2</-MX  inch;  b  = 
inch;  c  —  ^  </+2  inches;  £  =  3X  ^  +  3  inches;  ^= 
inch;       /=:#  ^+  X  inch  ;       ^-—1^^;      /=2^; 


Diameter  of  bolts,  //  =  0.2  ^/+  X  inch  (approximately). 
Diameter  of  bolts,  i  =  0.2  d  -f-  ^  inch  (approximately). 


SEARINGS. 


373 


FIG.   5. 


FIGURE  5  shows  a  bearing 
fitted  into  the  frame  of  a  ma- 
chine, suitable  for  shafts  from 
one  to  three  inches  in  diameter. 
The  cut  shows  a  part  of  the 
head-stock  of  a  speed  lathe 
fitted  with  this  kind  of  a  bear- 
ing. The  bearing  itself,  which 
may  be  of  gun  metal  or  cast 
iron,  is  carefully  fitted  into  the 
frame  by  planing  and  scrap- 
ing. 

This  kind  of  a  bearing  is 
sometimes  lined  with  babbitt, 
but  more  frequently  the  spindle 
is  carefully  fitted  into  the  bear- 
ing by  scraping. 


d  =  Diameter  of  bearing ;  /  =  2 d\  a=.2l4  d-\- 1  inch ; 
b  —  \y2d  +  %  inch;.*:  =  1#  d\  e  =  f  =  1%  d  +  %  inch  ; 
/=  e  =  lX<t+ys >inch;  S  =  %<*  +  %  inch;  £  =  !#</+# 
inch ;  i  =  2  d.  Diameter  of  screws  =  3A$  d  -f-  %e  inch. 

FIGURE  6  shows  the  form  of  a  self-lining  and  self-oiling 
bearing,  very  suitable  for  high-speed  machinery,  and  used  to  a 
great  extent  for  dynamos  and  electric  motors.  The  figure 
shows  a  part  of  a  dynamo  frame  with  the  box  in  section,  cut 
through  the  center  line  of  bearing,  and  also  a  partly  sectional 
cut  from  the  top.  For  dimensions  see  Table  No.  46. 

The  bearing  n  n  may  be  cast  in  one  piece  from  gun  metal,  as 
shown  in  the  cut,  or  it  may  be  (preferably  for  the  larger  size) 
made  in  two  parts.  The  seat  for  this  box  is  turned  in  spherical 
form  on  the  outside,  and  a  fit  is  obtained  between  this  bearing 
and  the  frame  of  the  machine  by  casting  in  type  metal  or  bab- 
bitt metal,  as  shown  at  m  m. 

The  loose  rings,  n  n,  are  continually  dipping  into  the 
oil  reservoir,  and  carrying  oil  to  the  shaft.  (Chains  are  fre- 
quently used  instead  of  rings).  The  stop-rings  should  be  set  so 
that  the  spindle  has  room  for  a  little  motion  lengthwise  in  the 
bearing.  This  will  in  a  great  measure  prevent  heating  and  cut- 
ting, and  by  their  peculiar  shape  the  stop-rings  will,  by  the 
action  of  centrifugal  force,  throw  the  oil  off  at  a  a,  to  return  to 
the  oil  reservoir;  h  h  are  plugs  in  the  oil  hole  ;  the  screw  i  pre- 
vents the  box  from  turning  with  the  shaft,  and  also  forms  a 
convenient  projection  to  take  hold  of  when  taking  the  cap  off 
of  the  bearing. 


374 


FIG. 


TABLE  No.  46.— Giving  Dimensions  of  Fig.  6. 


D 


Inches 


Inch. 


2/8 


234 
3 


5% 
5H 


87/8 

9, 
10 


125/Q 
13 


4 
5 

55/4 
6 

65/4 

7     • 


734 
8 


% 


H* 
6TV 


7/8 


10 


GEAR   TEETH.  375 

GEAR  TEETH. 

Circular  Pitch. 

The  length  of  the  pitch  circle  or  pitch  line  from  center  of 
one  tooth  to  the  center  of  the  next  is  the  circular  pitch  of  a  gear, 
or  a  rack. 

Cast  gear  teeth,  constructed  on  the  circular  pitch  system, 
may  be  made  of  the  following  proportions  : 

Thickness  of  tooth  on  pitch  line  =  T63-  pitch. 

Space  between  teeth  on  pitch  line  =  fj  pitch. 

Height  of  tooth  outside  of  the  pitch  line  =  T3ff  pitch. 

Depth  of  space  inside  of  pitch  line  =  T%  pitch. 

_,.    .    ..  circular  pitch  X  number  of  teeth. 

Pitch  diameter  of  gear  —  —  —  — 


For  cut  gears,  use  the  following  formulas  : 

Thickness  of  tooth  on  pitch  line  =  0.5  pitch. 
Space  between  teeth  on  pitch  line  =  0.5  pitch. 
Height  of  tooth  outside  pitch  line  —  0.3183  pitch. 
Depth  of  space  inside  of  pitch  line  =  0.3683  pitch. 

To  Calculate  Diameter  of  Gear  According  to  Circular  Pitch. 

RULE. 

Multiply  the  circular  pitch  by  the  number  of  teeth  in 
the  gear  and  divide  the  product  by  3.1416  ;  the  quotient  is 
the  diameter  of  the  pitch  circle  ;  add  T%  of  the  circular  pitch  to 
obtain  the  whole  diameter  of  the  gear. 

EXAMPLE. 

Find  whole  diameter  of  a  gear  of  48  teeth  and  three-inch 
circular  pitch. 

Solution  : 

Pitch  diameter  —  ^  *  *f  =  45.84  inches. 
3.1416 

Double  the  addendum  =  3  X  0.6  =  1.80 

The  whole  diameter  is  47.64  inches. 

Table  No.  47  is  calculated  for  one-inch  circular  pitch;  to 
find  the  pitch  diameter  of  a  gear  of  any  number  of  teeth  given 
in  the  table,  multiply  the  diameter  given  in  the  table  by  the 
circular  pitch  in  the  gear,  and  the  product  is  the  pitch  diameter 
of  the  gear.  In  order  to  find  the  whole  diameter,  add  twice  the 
height  of  the  tooth  outside  the  pitch  line,  as  calculated  by  the 
above  formula. 


376 


GfcAR  Tfcfcttf. 


TABLE  No.  47 — Giving  Pitch  Diameter  of  Gears  of  One 
Inch  Circular  Pitch. 


Teeth. 

Dia. 

Teeth. 

Dia. 

Teeth. 

Dia. 

Teeth. 

Dia. 

12 

3.82 

36 

11.46 

60 

19.10 

84 

26.74 

13 

4.14 

37 

11.78 

'  61 

19.42 

85 

27.06 

14 

4.46 

38 

12.10 

62 

19.74 

86 

27.38 

15 

4.78 

39 

12.42 

63 

20.06 

87 

27.70 

16 

5.09 

40 

12.73 

64 

20.37 

88 

28.01 

17 

5.41 

41 

13.05 

65 

20.69 

89 

28.33 

18 

5.73 

42 

13.37 

66 

21.01 

90 

28.65 

19 

6.05 

43 

13.69 

67 

21.33 

91 

28.97 

20 

6.37 

44 

14 

68 

21.65 

92 

29.29 

21 

6.69 

45 

14.32 

69 

21.97 

93 

29.60 

22 

7 

46 

14.64 

70 

22.28 

94 

29.92 

23 

7.32 

47 

14.96 

71 

22.60 

95 

30.24 

24 

7.64 

48 

15.28 

72 

22.92 

96 

30.56 

25 

7.96 

49 

15.60 

73 

23.24 

97 

30.88 

26 

8.28 

50 

15.92 

74 

23.56 

98 

31.20 

27 

8.60 

51 

16.24 

75 

23.88 

99 

31.52 

28 

8.91 

52 

16.55 

76 

24.19 

100 

31.83 

29 

9.23 

53 

16.87 

77 

24.51 

101 

32.15 

30 

9.55 

54 

17.19 

78 

24.83 

102 

32.47 

31 

9.87 

55 

17.51 

79 

25.15 

103 

32.78 

32 

10.19 

56 

17.83 

80 

25.47 

104 

33.10 

33 

10.50 

57 

18.14 

81 

25.79 

105 

33.42 

34 

10.82 

58 

18.46 

82 

26.10 

106 

33.74 

35 

11.14 

59 

18.78 

83 

26.42 

107 

34.06 

To  Calculate  Diameter  of  Gears   when  Distance  Between 
Centers  and  Ratio  of  Speed  is  Given. 

When  calculating  gears  to  connect  two  shafts  of  given  dis- 
tance between  centers  and  at  a  given  ratio  of  speed,  use  the 
formula, 

2  X  S  X  n 
n  +  N~ 
2X  SKN 
n  +  N 

D  =  Diameter  of  large  gear. 

*/—  Diameter  of  small  gear. 

S=  Distance  between  centers  in  inches. 

N=  Number  of  revolutions  of  large  gear  per  minute. 

n  =  Number  of  revolutions  of  small  gear  per  minute. 
NOTE.—  The  small  gear  is  always  on  the  shaft  having  the 
greater  speed. 


GEAR  TEETH.  377 

EXAMPLE. 

What  will  be  the  diameter  of  the  gears  to  connect  two  shafts 
when  the  distance  between  centers  is  32  inches,  and  one  shaft 
is  to  make  135  revolutions  and  the  other  105  revolutions  per 
minute  ? 

Solution : 

„       2  X  32  X  135 

L>  =  — 135  -L  iQ5 —  =  3"  inches  diameter. 

,       2  X  32   X  105 

135  4-  105 —  ~      mcnes  diameter. 

After  the  diameter  of  the  gears  is  calculated,  the  pitch  is 
decided  upon  according  to  the  power  the  gears  have  to  transmit. 

Frequently  the  pitch  will  have  to  be  altered  somewhat,  and 
such  gears  sometimes  have  teeth  of  very  odd  pitch,  in  order  to 
obtain  the  right  number  of  teeth  to  give  the  required  ratio  of 
speed.  The  ratio  between  the  number  of  teeth  in  the  gears  may 
always  be  seen  from  the  ratio  of  speed  between  the  two  shafts. 
For  instance,  in  the  above  example,  the  ratio  of  speed  between 
the  shafts  is  13%os,  which,  reduced  to  its  lowest  terms,  is  % ; 
therefore,  the  number  of  teeth  in  the  two  gears  may  be  any 
multiple  of  9  and  7,  respectively. 

For  instance,  8  X  9  =  72  teeth  for  the  large  gear,  and  8  X 
7  =  56  teeth  for  the  small  gear ;  or,  10  X  9  =  90  teeth  for  the 
large  gear,  and  10  X  7  =  70  teeth  for  the  small  gear,  etc. 

The  dimensions  of  teeth  may  be  calculated  according  to 
rules  given  on  page  375. 

Diametral  Pitch. 

The  diametral  pitch  of  a  gear  is  the  number  of  teeth 
to  each  inch  of  its  pitch  diameter.  In  cut  gearing  it  is  always 
customary  to  calculate  the  gears  according  to  diametral  pitch. 
When  gears  are  calculated  according  to  circular  pitch  the  corre- 
sponding circumference  of  the  pitch  circle  is  usually  an  even 
number,  but  the  diameter  will  generally  be  a  number  having 
cumbersome  fractions,  and  therefore  the  distance  between  the 
centers  of  the  gears  will  be  a  number  having  fractions  which 
may  be  very  inconvenient  to  measure  with  common  scales. 
This  is  because  the  circumference  of  a  circle  divided  by 
3.1416  is  equal  to  its  diameter  and  the  diameter  multi- 
plied by  3.1416  is  equal  to  the  circumference.  When 
gearing  is  calculated  according  to  diametral  pitch  this  trouble 
is  entirely  avoided,  as  this  directly  expresses  the  number  of 
teeth  on  the  circumference  of  the  gear  according  to  its  pitch 
diameter.  For  instance,  "six  diametral  pitch"  means  that 
there  are  six  teeth  on  the  circumference  of  the  gear  for  each 
inch  of  pitch  diameter.  Thus,  a  gear  of  six  diametral  pitch  and 
forty-eight  teeth  will  be  eight  inches  pitch  diameter.  A  gear  of 
"  eight  diametral  pitch  "  means  that  the  gear  has  eight  teeth  per 


378  GEAR   TEETH. 

inch  of  pitch  diameter.  A  gear  of  "  ten  diametral  pitch  "  means 
that  the  gear  has  ten  teeth  per  inch  of  pitch  diameter.  A  gear 
of  "  twelve  diametral  pitch  "  means  that  the  gear  has  twelve 
teeth  per  inch  of  pitch  diameter,  etc. 

Thus,  the  pitch  diameter  and,  consequently,  the  distance 
between  the  centers,  will  be  a  number  which  may  be  conven- 
iently measured,  and  the  dimensions  of  tooth  parts  are  also 
much  more  easily  calculated  by  this  system. 

Rules  for  Calculating  Dimensions  of  Gears  According  to 
Diametral  Pitch. 

The  pitch  diameter  is  obtained  by  dividing  the  number  of 
teeth  by  the  diametral  pitch. 

EXAMPLE. 

What  is  the  pitch  diameter  of  a  gear  of  48  teeth,  16  pitch? 
Solution : 

48  divided  by  16  =  3,  therefore  the  pitch  diameter  is  3 
inches. 

The  number  of  teeth  is  obtained  by  multiplying  the  pitch 
diameter  by  the  diametral  pitch. 

EXAMPLE. 

What  is  the  number  of  teeth  in  a  gear  of  5  inches  pitch 
diameter  and  12  pitch  ? 

Solution : 

5  X  12  =  60,  therefore  the  gear  has  60  teeth. 

The  whole  diameter  of  a  spur  gear  is  obtained  by  adding 
2  to  the  number  of  teeth  and  dividing  the  sum  by  the  diametral 
pitch. 

EXAMPLE. 

What  is  the  whole  diameter  of  a  gear  blank  for  68  teeth, 
10  pitch  ? 

Solution : 

68    I    2 
Whole  diameter  =  — :— —  =  7  inches. 

The  number  of  teeth  is  obtained  by  multiplying  the  whole 
diameter  of  the  gear  by  the  diametral  pitch  and  subtracting  2 
from  the  product. 

EXAMPLE. 

The  whole  diameter  of  a  gear  blank  is  8  inches ;  it  is  to 
be  cut  10  diametral  pitch.  Find  the  number  of  teeth. 

Solution : 

Number  of  teeth  =  (8  X  10)  —  2  =  78. 

The  diametral  pitch  is  obtained  by  adding  2  to  the  number 
of  teeth  and  dividing  by  the  whole  diameter. 


GEAR  TEETH.  379 

EXAMPLE. 

A  gear  has  64  teeth  and  the  whole  diameter  is  \Ql/2  inches. 
What  is  the  diametral  pitch  ? 
Solution  : 

Diametral  pitch  =       —  r-  =  4. 


Thus,  the  gear  is  4  diametral  pitch. 

NOTE.  —  The  term  diameter  of  a  gear  usually  means 
ter  of  pitch  circle. 

The  distance  between  the  centers  of  two  spur  gears  is  ob- 
tained by  dividing  half  the  sum  of  their  teeth  by  the  diametral 
pitch. 

EXAMPLE. 

What  is  the  distance  between  centers  of  two  gears  of  48 
and  64  teeth  and  8  diametral  pitch  ? 
Solution  : 

Distance  =  4f  j"^4  =  7  inches. 

2i  X  o 

The  circular  pitch  is  obtained  by  dividing  the  constant 
3.1416  by  the  diametral  pitch. 
EXAMPLE. 

What  is  the  circular  pitch  of  a  gear  of  eight  diametral 
pitch  ? 

Solution  : 


Circular  pitch  =  .  —  0.393  inch. 

o 

The  thickness  of  the  tooth  on  the  pitch  line  is  obtained  by 
dividing  the  constant  1.5708  by  the  diametral  pitch. 
EXAMPLE. 

What  is  the  thickness  of  the  tooth  on  the  pitch  line  of  a 
gear  of  6  diametral  pitch  ? 
Solution  : 


1   P 

Thickness  of  tooth  =  ^—  ,  —  =  0.262  inches. 
6 

The  working  depth  of  the  tooth  is  obtained  by  dividing  2 
by  the  diametral  pitch.  The  clearance  at  the  bottom  of  the 
teeth  is  TV  of  the  thickness  of  the  tooth  on  the  pitch  line.  The 
whole  depth  to  cut  the  gear  is  obtained  by  dividing  the  constant 
2.157  by  the  diametral  pitch. 

EXAMPLE. 

Find  the  depth  to  cut  a  gear  of  8  diametral  pitch. 
Solution  : 

Depth  =  2-*37-  =  0.27  inch. 

o 


380 


GEAR  TEETH. 


The  whole  depth  is  nearly  equal  to  0.0806  times  the  circular 
pitch.  The  use  of  the  following  tables  will  facilitate  calcula- 
tions regarding  dimensions  of  teeth  in  diametral  pitch. 


TABLE  No.  48.— Comparing  Circular  and  Diametral  Pitch. 


Diametral  Pitch. 

Circular  Pitch. 

Circular  Pitch. 

Diametral  Pitch. 

2 

1.571  inch. 

ll/2  inch. 

2.094 

2^ 

1.257     " 

1A 

2.185 

3 

1.047     " 

2.285 

3^ 

0.898     " 

1A 

2.394 

4 

0.785     " 

IX 

2.513 

5 

'  0.628     " 

1  3 

2.646 

6 

0.524     " 

i/^ 

2.793 

7 

0.449     " 

1A 

2.957 

8 

0.393     " 

i 

3.142 

9 

0.349     " 

if 

3.351 

10 

0.314     " 

?i 

3.590 

11 

0.286     " 

il 

3.867 

12 

0.262     " 

X 

4.189 

14 

0.224     " 

4.570 

16 

0.196     " 

$& 

5.027 

18 

0.175     " 

_9 

5.585 

20 

0.157     " 

J& 

6.283 

22 

0.143     " 

A 

7.181 

24 

0.131     " 

y% 

8.378 

26 

0.121     " 

A 

10.053 

28 

0.112     " 

V      ' 

12.566 

30 

0.105     " 

fV    " 

16.755 

32 

0.098     " 

r/         U 

25.133 

TABLE  No.  49.—  Giving  Dimensions  of  Teeth  Calculated 

According  to  Diametral  Pitch. 

Diametral 
Pitch. 

Depth  to  be 
Cut  in  Gear. 

Thickness  of 
Tooth  on 
Pitch  Line. 

Diametral 
Pitch. 

Depth  to  be 
Cut  in  Gear. 

Thickness  of 
Tooth  on 
Pitch  Line. 

2 

1.078  in. 

0.785  in. 

12 

0.180  in. 

0.131  in. 

2/^ 

0.863 

0.628 

14 

0.154 

0.112 

3 

0.719 

0.523 

16 

0.135 

0.098 

3/2 

0.616 

0.448 

18 

0.120 

0.087 

4 

0.539 

0.393 

20 

0.108 

0.079 

5 

0.431 

0.314 

22 

0.098 

0.071 

6 

0.359 

0.262 

24 

0.090 

0.065 

7 

0.307 

0.224 

26 

0.083 

0.060 

8 

0.270 

0.196 

28 

0.077 

0.056 

9 

0.240 

0.175 

30 

0.072 

0.052 

10 

0.216 

0.157 

32 

0.067 

0.049 

11 

0.196 

0.143 

GEAR   TEE 


To  Calculate  the  Number  of  Teeth   when   Distance  Be- 
tween Centers  and  Ratio  of  Speed  is  Given. 

Select  for  a  trial  calculation,  the  diametral  pitch  which 
seems  most  suitable  for  the  work. 

Calculate  the  sum  of  the  number  of  teeth  in  both  gears 
corresponding  to  this  pitch  by  multiplying  twice  the  distance 
between  their  centers  by  the  diametral  pitch  selected. 

The  number  of  teeth  in  each  gear  is  obtained  by  the  follow- 
ing formula : 


._  n  X  A 
i  — 


N+n 

T=  Number  of  teeth  in  large  gear. 

/  =  Number  of  teeth  in  small  gear. 
N=.  Number  of  revolutions  of  small  gear. 
n  =  Number  of  revolutions  of  large  gear. 
A  =  Number  of  teeth  in  both  gears. 

EXAMPLE. 

The  center  distance  between  two  shafts  is  15  inches.  The 
small  gear  should  make  126  and  the  large  gear,  90  revolutions 
per  minute.  Calculate  the  number  of  teeth  in  each  gear,  if  8 
diametral  pitch  is  wanted. 

Solution  : 

The  number  of  teeth  in  both  gears  is  2  X  15  X  8  =  240. 


126  +  90 

t  =   90  x  24°  =  100  teeth. 
126  +  90 

Frequently  it  is  impossible  to  get  gears  of  the  desired  pitch 
to  fit  within  the  given  center  distance  and  to  give  the  exact 
ratio  of  speed.  Some  modifications  must  then  be  made  ;  either 
the  exact  ratio  of  speed  must  be  sacrificed,  the  pitch  must  be 
changed,  or  the  distance  between  centers  must  be  altered. 

NOTE.  —  The  ratio  of  the  number  of  teeth  in  the  gears  can  be 
seen  from  the  ratio  of  the  speed.  For  instance,  in  the  above  ex- 
ample the  ratio  of  speed  is  °%26,  which,  reduced  to  its  lowest 
terms,  is  %  ;  therefore,  the  number  of  teeth  in  the  two  gears 
may,  with  regard  to  speed  ratio,  be  any  multiple  of  5  and  7, 
respectively,  but  in  order  to  fit  the  given  center  distance  and 
also  to  be  8  pitch,  they  must  be  100  and  140,  which  is  20  X  $ 
=  100  and  20  X  7  =  140. 


•82 


GEAR    TEETH. 


FIG.  1 


The  shape  of  gear  teeth  is  usually  either  Involute  or  Cycloid 
(  also  frequently  called  Epicycloid ).  The  shape  of  a  cycloid 
looth  for  a  rack  is  four  equal  cycloid  curves,  which  may  be  con- 
structed, so  to  speak,  by  letting  the  generating  circle  a,  ( sec 
Fig.  1 )  roll  along  on  the  pitch  line  of  the  rack,  both  above  and 
below.  / 

Cycloid  gears 
have  the  curve  out- 
side the  pitch  circle 
formed  by  an  Epi- 
cycloid (see  Fig.  20, 
page  191)  and  the 
curve  inside  the 
pitch  circle  by  a 
Hypocycloid. 

The  curves  al- 
ways meet  on  the 
pitch  line  in  both 
gears  and  racks. 

The  theoretical 
requirements  for 
correct  form  of  Epi- 
cycloid gear  teeth 
are  that  the  face  of 
the  teeth  of  one  gear  and  the  flank  of  the  teeth  of  the  other  gear 
must  be  produced  by  generating  circles  of  the  same  diameter. 

The  diameter  of  the  generating  circle  is  limited  by  the  size 
of  the  smallest  gear  or  pinion  in  the  series  of  gears  which  are  con- 
structed to  run  together,  because  if  the  generating  circle  is  as 
large  in  diameter  as  half  the  pitch  diameter  of  the  gear,  the 
hypocycloid  will  be  a  straight  line ;  thus,  the  flank  of  the  tooth 
will  be  a  straight  radial  line.  If  the  generating  circle  is 
larger  than  half  the  pitch  diameter  of  the  gear,  the  result  will  be 
a  weak  and  poor  tooth  with  under-cut  flank. 

When  the  same  size  of  generating  circle  is  used  for  gears  of 
different  diameters  but  of  the  same  pitch,  all  such  gears  will  work 
correctly  together,  and  for  this  reason  it  is  possible  to  construct 
interchangeable  gears  having  cycloid  teeth.  If  the  diameter  of 
the  generating  circle  is  equal  to  half  the  diameter  of  the 
smallest  gear  in  the  set,  this  gear  will  have  teeth  with  radial 
flanks  but  all  the  other  gears  and  the  rack  will  have  double- 
curved  teeth.  Fig.  1  shows  a  rack  drawn  to  >£-inch  circular 
pitch  ;  the  generating  circle  is  0.98  inch  diameter,  which  is  equal 
to  half  of  the  pitch  diameter  of  a  gear  of  12  teeth  and  y2  inch 
circular  pitch. 

All  gears  of  the  same  pitch  having  12  teeth  or  more,  con- 
structed by  the  same  generating  circle  in  the  same  manner  as 
the  rack,  will  match  and  be  interchangeable  with  the  rack,  and 
will  also  match  and  be  interchangeable  with  each  other. 


GEAR   TEETH.  383 

When  internal  teeth  are  constructed  by  the  above  method,  the 
difference  between  the  number  of  teeth  in  the  internal  gear  and 
its  external  pinion  must  never  be  less  than  12;  practically,  it  is 
better  to  limit  the  difference  to  15  or  20  teeth. 

As  interchangeability  is  seldom  required  for  internal  gear- 
ing, such  gears  and  their  mates  are  generally  constructed  together 
and  the  designer  chooses  a  generating  circle  of  suitable  size  to 
give  the  shape  of  tooth  he  considers  best,  and  he  may  also  vary 
the  size  of  the  driving  or  the  driven  gear  so  as  to  reduce  con- 
tacts when  the  teeth  are  approaching  each  other,  etc.,  according 
to  his  own  judgment  and  experience. 

The  difference  in  pitch  diameter  of  the  internal  gear  and  its 
pinion  should  never  be  less  than  the  sum  of  the  diameters  of  the 
generating  circles,  and  the  diameter  of  the  generating  circle  of 
the  flanks  for  the  pinion  should  never  be  larger  than  half  the 
pitch  diameter,  but  it  should,  preferably,  be  smaller. 

As  a  rule,  fillets  at  the  bottom  of  the  teeth  are  not  used  in 
internal  gears,  but  if  used  they  should  be  very  small. 

In  order  that  gears  constructed  with  cycloid  teeth  should 
run  smoothly,  it  is  very  important  to  have  the  distance  between 
centers  correct,  so  that  the  pitch  lines  will  exactly  meet  each 
other.  For  this  reason,  there  are  many  kinds  of  machinery 
where  cycloid  gears  should  not  be  used :  for  instance,  for  change 
gears  on  lathes,  involute  teeth  as  far  more  suitable. 

When  making  patterns,  the  shape  of  one  tooth  is  usually 
carefully  drawn  on  a  thin  piece  of  sheet  metal,  either  brass  or 
iron ;  this  is  then  filed  out  and  used  as  a  templet  in  tracing 
the  other  teeth  on  the  pattern.  Sometimes  a  fly-cutter  is  made 
according  to  this  constructed  tooth,  and  all  the  teeth  in  the  pat- 
tern are  cut  on  an  index  machine  or  a  gear  cutting  machine ;  but 
if  such. a  machine  is  not  available,  the  next  best  way  is  to  set  out 
the  pitch  line  of  the  gear  on  this  templet  and  also  the  center  line 
of  the  tooth,  radially  towards  the  center,  then  draw  the  pitch  line 
on  the  pattern,  space  off  each  tooth  carefully  with  a  pair  of 
dividers  and  draw  the  center  line  on  each  tooth  prolonged  across 
the  rim  radially  in  the  direction  of  the  center  of  the  gear,  then 
lay  the  templet  carefully  on  each  of  these  spacings,  making 
the  pitch  line  and  the  center  line  of  tooth  on  the  templet  to 
exactly  match  the  pitch  line  and  center  line  of  the  tooth 
drawn  on  the  pattern,  then  trace  around  the  templet  and  get  the 
shape  of  one  tooth ;  then  move  the  templet  to  the  next  spacing 
and  trace  the  next  tooth,  and  so  on  for  all  the  teeth  on  the  gear. 

For  small  patterns  it  is  convenient  to  fasten  the  templet  to 
a  strip  of  metal  long  enough  to  reach  from  the  teeth  to  the  cen- 
ter of  the  gear  wheel,  placing  a  point  in  the  center  of  the  gear, 
drilling  a  hole  in  the  strip  and  letting  it  swing  around  this  point, 
then  after  all  the  teeth  are  spaced  off  on  the  pattern  the  tem- 
piet  is  swung  from  one  tooth  to  the  other  and  all  the  teeth  are 
traced  by  the  templet.  This  method  has  the  advantage  that 


384  GEAR   TEETH. 

it  will  mark  all  the  teeth  exactly  alike,  because  ,the  templet 
being  fastened  to  this  strip,  can  not  easily  get  out  of  position. 

The  distance  from  the  pitch  line  of  the  templet  to  the  cen- 
ter hole  in  the  strip  must  be  laid  off  according  to  the  shrinkage 
rule,  and  is,  of  course,  in  numerical  value  equal  to  the  pitch  radius 
of  the  gear,  which  should  always  be  calculated  and  given  on  the 
drawing.  When  gear  patterns  are  less  than  six  inches  in  diame- 
ter it  is  preferable  not  to  allow  anything  for  shrinkage,  as  the 
moulder  will  usually  rap  the  pattern  about  as  much  as  the  cast- 
ing will  shrink  in  cooling. 

When  a  pair  of  cycloid  gears  are  constructed  without  con- 
sidering interchangeability  with  other  gears  of  the  same  pitch, 
it  is  customary  to  choose  a  generating  circle  having  a  diameter 
equal  to  three-fourths  of  the  radius  of  the  pitch  circle  of  the 
small  gear,  providing  this  gear  has  24  or  more  teeth.  A  large 
generating  circle  probably  reduces  the  friction  in  a  small 
measure  but  gives  teeth  of  less  strength.  The  largest  gen- 
erating circle  used  ought  never  to  exceed  the  radius  of  the 
pitch  circle  of  the  small  gear.  Decreasing  the  generating 
circle  will  probably  increase  friction  somewhat  in  the  gears,  but 
it  gives  teeth  of  greater  strength.  The  smallest  generating  circle 
used  in  practice  is  equal  to  half  the  diameter  of  the  pitch  circle 
of  a  gear  having  12  teeth  of  the  same  pitch  as  the  gear  to  be  con- 
structed. Many  eminent  mechanics  consider  it  preferable 
never  to  use  a  generating  circle  smaller  than  half  of  the  pitch 
diameter  of  a  gear  of  15  teeth. 

Cycloid  gears  are  mostly  used  in  large  cast  gears  of  one- 
inch  circular  pitch  or  more. 

Sometimes  the  driving  gear  is  made  of  slightly  larger  di- 
ameter, and  the  teeth  spaced  at  a  correspondingly  greater  pitch 
than  the  theoretically  calculated  size.  This  is  done  in  order 
that  the  teeth  shall  not  rub  on  each  other  on  the  approaching 
side,  but  only  touch  as  they  are  passing  the  center  line  and 
commence  to  slide  away  from  each  other.  This  will  make  the 
gears  less  noisy,  but  probably  gears  made  in  this  manner  will 
wear  faster,  as  there  are  fewer  points  of  contact,  although  this 
may  be  offset  by  the  fact  that  the  friction  between  the  teeth 
when  they  are  meeting  and  pushing  onto  each  other  is  more 
injurious  than  the  friction  produced  when  they  are  sliding  away 
from  each  other. 

The  same  idea  is  sometimes  employed  when  constructing 
bevel  gears,  in  order  to  make  them  run  quietly. 

This  mode  of  sizing  gears  is  not,  as  a  rule,  used  in  modern 
gear  construction,  but  it  is  a  point  well  worth  remembering, 
because  if  either  of  two  gears  is  over  or  under  size,  the  gear  of 
over-size  should  always  be  used  as  the  driver,  and  the  gear  of 
under-size  should  always  be  the  driven ;  never  vice  versa.  This 
will  apply  as  well  to  involute  as  to  epicycloid  gears. 


GEAR   TEETH. 


3*5 


Involute  Teeth. 

Suppose  a  strap  is  fastened  at  a  and  b  on  the  two  round  discs 
in  Fig.  2.  If  the  disc  b  is  turned  in  the 
direction  of  the  arrow,  the  strap  will  move 
in  a  straight  line  from  c,  toward  d.  This 
motion  will  cause  the  disc  a  to  rotate  with 
exactly  the  same  surface  speed  as  the  disc 
b,  but  in  the  opposite  direction. 

Suppose,  further,  that  to  the  under  side 
of  the  disc  a  (see  Fig.  3)  is  fastened  a  piece 
of  sheet  brass  p,  or  other  suitable  material 
of  somewhat  larger  diameter  than  disc  a, 
and  that  a  scratch  awl  is  fastened  in  the 
strap  at  the  point  /// ;  then  by  turning  the 
disc  b  in  the  direction  of  h  to  b,  and 
the  strap  moving  with  it,  being  kept  tight 
by  the  resistance  of  disc  a,  the  scratch  awl 
will  trace  on  the  brass  plate  the  curve  from 
in  to  h,  but  if  the  discs  are  moving  in  the 
opposite  direction,  the  scratch  awl  will 
trace  the  curve  from  ;//  to  K,  Take  an- 
other brass  plate  and  do  the  same  thing  with  the  other  disc,  and 
a  similar  curve  will  be  produced.  In  these  two  brass  plates  the 
stock  may  be  filed  away  carefully,  following  the  curves  as  shown 
in  Fig.  4.  The  discs  are  laid  to  match  each  other  and  free  to 


FIG.  2. 


FIG.  4. 


FIG.  3. 


swing  on  their  centers ;  turning  the  disc  a  in  the  direction  of  the 
arrow,  it  will  give  motion  to  b,  and  both  discs  will  move  with  the 
same  speed  in  exactly  the  same  manner  as  if  they  were  connected 
by  the  strap  as  shown  in  Fig.  2. 


386 


GEAR  TEETH. 


The  curve  on  these  two  discs  represents  the  form  of  a  gear 
tooth  in  the  involute  system.* 

The  line  h  g,  Fig.  4,  is  called  the  line  of  pressure  or  the  line 
of  action.  The  circles,  P  and  P,  are  the  pitch  circles.  The  line 
B  R,  shows  the  direction  of  motion  of  the  teeth  at  the  moment 
they  are  passing  the  center  line,  c  c. 


Approximate  Construction  of  Involute  Teeth. 

It  will  be  noticed  that  the  line  of  pressure,  h  g,  forms  an 
angle  with  the  line  B  R.  This  angle  is  usually  taken  as  14)4 
degrees.  This  makes  the  diameter  of  the  base  circle,  ^,  (see 
Fig.  5)  equal  to  0.968  times  the  diameter  of  the  pitch  circle. 
The  base  circle  g  g,  in  Fig.  5,  corresponds  to  the  disc  in  Fig.  3, 
and  the  line  of  pressure  in  Figs.  5  and  6  corresponds  to  the  strap 
in  Fig.  2.  The  line  of  pressure,  h  g,  Fig.  5,  is  75 >£  degrees  to 
the  center  line,/Y. 

A  perpendicu- 
lar is  erected 
from  the  line  h 
g,  through  the 
center,  c.  Using 
the  point  of  in- 
tersection at  i  as 
center,  the  tooth 
is  drawn  simply 
by  a  circular  arc. 
This  will,  in  prac- 
tical work  for 
small  gears  hav- 
ing more  than 
twenty  teeth,  cor- 
respond nearly 
enough  to  the 
true  involute,  which  was  illustrated  by  means  of  the  strap,  disc 
and  scratch  awl,  as  explained  in  Figs.  2,  3  and  4. 

When  the  gear  has  less  than  twenty  teeth,  and  is  constructed 
by  circular  arcs,  as  shown  in  Fig.  5,  the  top  of  the  tooth  will  be 
too  thin ;  but  the  top  of  the  tooth  will  be  too  thick  to  clear  in 
the  rack,  if  the  true  involute  curve  is  used. 

When  the  teeth  are  of  true  involute  curve,  a  smaller  gear 
than  twenty-five  teeth  will  not  run  freely  in  a  rack  having  straight 
teeth  slanting  14^  degrees.  (See  Figs.  6  and  7).  Therefore, 


*The  way  to  actually  draw  this  curve  on  paper  by  means  of  drawing  instru- 
ments is  explained  on  page  192.  This  way  explained  here,  using  the  disc  on  the 
strap,  is  merely  for  illustrating  and  explaining  principles,  and  serves  well  for  that 
purpose,  but  would  be  inconvenient  to  use  in  actual  construction  of  gear  teeth.  In 
actual  work  one  tooth  is  carefully  constructed,  and  templets  and  cutters  are  made 
and  used,  as  was  explained  for  Cycloid  Gears,  pa^e  383. 


GEAR  TEETH. 


when  a  gear  has  less  than  twenty-five  teeth  it  is  necessary  to  round 
the  teeth  somewhat  outside  the  pitch  circle.  By  making  either  a 
drawing  or  a  templet,  it  is  very  easy  to  see  how  much  to  round 


FIG.  6.    Involute  Teeth  (Cast.) 


the  teeth  to  make  them  clear  in  the  rack.  In  interchangeable 
sets  of  cut  involute  gears  it  is  customary  to  cut  the  rack  with  a 
cutter  shaped  for  a  gear  of  185  teeth.  This  will  make  the  teeth 
in  the  rack  slightly  curved  instead  of  straight,  as  shown  in  Fig.  6, 
and  this  will  also  make  it  possible  to  construct  the  small  gears  in 
an  interchangeable  set  nearer  to  a  true  involute,  and  still  have 
them  run  freely  in  the  rack. 


388 


GEAR   TEETH. 


When  gear  teeth  are  constructed  as  shown  in  Figs.  6  and  7, 
the  line  g  h,  is  75^  degrees  to  the  line  c  f,  and  the  line  c  /,  is 
degrees  to  the  line  c  f.     (See  Fig.  5). 


FIG.  7.    Involute  Teeth  (Cut). 

The  line  h  g,  will  always  be  tangent  to  the  base  circle 
which  is  concentric  to  the  pitch  circle.  The  diameter  of  the 
base  circle  is  always  0.9G8  times  the  diameter  of  the  pitch  circle. 
The  circle  forming  the  shape  of  the  tooth  must  always  have  its 
center  on  the  circumference  of  the  base  circle,  and  its  diameter 
will  be  one-fourth  of  the  pitch  diameter  of  the  gear.  As  shown 
in  Fig.  2,  the  same  circle  gives  the  form  of  tooth  for  coarser  or 
finer  pitch.  When  gears  are  drawn  by  this  method  the  pitch 
circle  is  divided  into  as  many  teeth  and  spaces  as  there  are  to 
be  teeth  in  the  gear ;  then  the  form  of  the  tooth  is  simply  struck 
by  the  dividers,  always  using  the  periphery  of  the  base  circle  as 
center,  and  always  taking  the  distance  in  the  dividers  equal  to 
one-fourth  of  the  radius  of  the  pitch  circle. 

The  diameter  of  the  base  circle  is  0.968  times  the  diameter  of 
the  pitch  circle,  because  cosine  of  14^  degrees  is  0.96815.  The 
diameter  of  the  circle  forming  the  shape  of  the  tooth  is  0.25  times 
the  diameter  of  the  pitch  circle  of  the  gear,  because  sine  of  14^ 
degrees  is  0.25038.  If  the  line  of  pressure  is  laid  at  any  other  angle 


GEAR   TEETH.  389 

than  14^  degrees,  all  these  other  proportions  will  also  change. 
Fig.  0  shows  a  pattern  for  gears  and  rack  constructed  with 
necessary  clearance  as  used  for  cast  gears.  All  tooth  parts  are 
of  the  same  dimensions  as  used  for  cycloid  gears  as  given  on 
page  375.  Fig.  7  shows  a  cut  gear  and  rack  constructed  in  the 
same  manner.  The  advantages  of  the  involute  system  of  gears 
are  in  the  strength  of  teeth,  and  also  that  the  gears  will  trans- 
mit uniform  motion  and  run  satisfactorily,  even  if  the  distance 
between  centers  should  be  slightly  incorrect. 

Width  of  Gear  Wheels. 

Gears  with  cast  teeth  are  usually  made  narrower  than  gears 
with  cut  teeth.  In  spur  gears  with  cast  teeth  it  is  customary  to 
make  the  width  of  the  gear  four  to  five  times  the  thickness  of 
the  teeth,  or  twice  the  circular  pitch. 

Width  of  Gears  With  Cut  Teeth. 

The  following  rule  is  recommended  by  Brown  &  Sharpe 
Mfg.  Co.  in  their  "  Practical  Treatise  on  Gearing": 

Divide  eight  by  the  diametral  pitch,  and  add  one-fourth  inch 
to  the  quotient;  the  sum  will  be  the  width  of  face  for  the  pitch 
required. 

EXAMPLE. 

What  width  of  face  is  required  for  a  gear  of  four  pitch  ? 

Solution  : 

Face  =  f  +  ]-  =  2%  inches. 

For  change  gears  on  lathes  where  it  is  desirable  not  to  have 
faces  very  wide,  the  following  rule  may  be  used : 

Divide  four  by  the  diametral  pitch  and  add  one-half  inch. 

By  the  latter  rule  a  four-pitch  change  gear  would  have  but 
a  1^-inch  face. 


BEVEL  GEARS. 

Fig.  8  is  a  diagram  showing  how  to  size  bevel  gear  blanks. 

First,  lay  off  the  pitch  diameters  of  the  two  gears,  which 
may  be  calculated  according  to  diametral  pitch  or  to  circular 
pitch ;  second,  draw  the  pitch  line  of  teeth ;  third,  lay  off  on  the 
back  of  the  gear  the  line  a  b,  square  to  the  "  pitch  line  of  teeth  ;" 
fourth,  on  the  line  a  b,  lay  off  the  dimensions  of  the  teeth  exactly 
in  the  same  manner  as  if  it  was  for  a  spur  gear. 

If  the  gear  is  calculated  according  to  circular  pitch,  find 
dimensions  of  teeth  by  formulas  on  page  375,  but  if  tne  gear  is 
calculated  according  to  diametral  pitch,  find  dimensions  of  teeth 
in  Table  No.  49. 


39° 


feEVEL  GEARS. 


Make  the  drawing  carefully  to  scale  ( full  size  preferable 
whenever  possible ),  and  measure  the  outside  diameter  as  shown 
in  the  diagram. 


FIG.  8. 


To  Calculate  Size  of  Bevel  Gear  for  a  Given 
Ratio  of  Speed. 

Ascertain  the  ratio  of  speed  in  its  lowest  terms.  Multiply 
each  term  separately  by  the  same  number,  and  the  products 
give  the  number  of  teeth  in  each  gear. 

EXAMPLE. 

Two  shafts  are  to  be  connected  by  bevel  gears,  one  shaft  to 
make  80  revolutions  and  the  other  170  revolutions  per  minute. 
Find  the  number  of  teeth  in  the  gears. 


BEVEL  GEARS. 


391 


Solution : 


For 


Ratio  = 


=  8/i7. 


instance,  multiplying  by  6,  the  large  gear  on  the  shaft 
making  80  revolutions  will  have  17X6  =  102  teeth.  The  small 
gear  on  the  shaft  making  170  revolutions  will  have  8  X  6  =  48 
teeth. 

Assuming  that  on  account  of  room  it  is  necessary  to  use 
smaller  gears,  a  smaller  multiplier  may  be  used,  but  if  it  is  desir- 
able to  have  larger  gears,  use  a  larger  multiplier. 

Decide  on  the  pitch  of  the  gears  according  to  the  work  they 
are  required  to  do.  Make  a  scale  drawing  and  get  the 
dimensions  as  explained  on  page  390. 

Dimensions  of  Tooth  Parts  in  Bevel  Gears. 

Fig.  9  shows  a  sectional  drawing  of  a  pair  of  bevel  gears  of 
sixteen  diametral  pitch,  18  teeth  in  the  small  gear  and  30  teeth 
in  the  large  gear.  The  pitch  diameter  of  the  small  gear  is  -££  = 
iVs  inches.  The  pitch  diameter  of  the  large  gear  is  f f  =  1% 
inches. 

FIG.  9. 


The  addendum  of  the  teeth  on  the  back  at  a  is  TV  inch, 
the  same  as  for  a  spur  gear  of  16  diametral  pitch.     The  thickness 


392  BEVEL   GEARS. 

and  the  total  depth  to  cut  the  gear  at  a  are  0.098  inch  and  0.135 
inch,  respectively. 

These  dimensions  are  found  in  Table  No.  40,  as  if  it  was 
a  spur  gear  of  10  diametral  pitch.  All  the  dimensions  of  the 
tooth  decrease  gradually  toward  b,  as  the  whole  tooth  is  sup- 
posed to  vanish  in  a  point  in  the  center  at  c.  The  dimensions 
of  the  teeth  at  b  may  be  calculated  and  are  always  in  tke  same 
proportion  to  the  dimensions  at  a  as  the  distance  c  b  is  to  the 
distance  c  a  ;  thus,  if  the  length  of  the  tooth  from  a  to  b  is  made 
one-third  of  the  length  of  the  distance  c  X  the  distance  b  c  is 
two-thirds  of  the  distance  a  c,  and,  consequently,  all  the  dimen- 
sions of  the  tooth  at  b  are  two-thirds  of  the  dimensions  at  a. 
Instead  of  calculating  the  size  of  the  teeth  at  b,  the  dimensions 
may  be  obtained  by  careful  drawing.  The  depth  of  the  tooth  at 
the  smallest  end  is  then  measured  directly  at  b,  but  the  thickness 
is  measured  at  /;  the  distance  /  h  is  laid  off  equal  to  b  d. 

The  length  of  the  tooth  from  a  to  b  is  to  a  certain  extent 
arbitrary,  but  a  good  rule  is  seven  inches  divided  by  the  diame- 
tral pitch,  but  never  longer  than  one-third  of  the  distance  from 
a  to  c. 

EXAMPLE. 

What  is  the  proper  length  for  the  teeth  of  a  bevel  gear  of  8 
diametral  pitch  ? 

Solution : 

Seven  inches  divided  by  8  —  %  inch,  if  the  gears  are  of  such 
diameters  that  this  will  not  make  the  length  of  the  teeth  more 
than  one-third  of  the  distance  from  a  to  c. 


Form  of  Tooth  in  Bevel  Gears. 

Extend  the  line  a  (see  Fig.  9),  until  it  intersects  the  axial 
center  line  of  the  gear,  as  at  h  ;  use  h  as  the  center,  and  the  shape 
of  tooth  at  a  for  the  large  gear  is  constructed  as  if  it  was  a  spur 
gear  having  a  pitch  radius  as  large  as  a  h. 

The  shape  of  the  tooth  at  b  is  constructed  in  the  same  way, 
by  extending  the  line  ^,  (which  always — the  same  as  line  a, — is 
square  to  the  pitch  line  of  the  tooth)  until  it  intersects:  the 
axial  center  line  of  the  gear,  as  at  d.  Using  d  as  center,  the 
shape  of  the  tooth  is  constructed  as  if  it  was  a  spur  gear  having 
a  pitch  radius  equal  to  db.  The  shape  of  the  teeth  of  the  small 
gear  is  obtained  in  the  same  way,  which  is  shown  by  the  drawing. 

The  form  of  tooth  is  shown  to  be  approximately  involute, 
constructed  as  explained  for  spur  gears,  page  386. 

Measuring  the  back  cone  radius,  a  h,  of  the  large  gear,  it  is 
found  to  be  f  f  inch,  and  the  diameter  will  be  f- 1  inch ;  thus,  the 
shape  of  the  tooth  at  a  for  the  large  gear  will  be  the  same  as  the 
shape  of  the  tooth  in  a  spur  gear  of  58  teeth,  sixteen  diametral 
pitch. 


BEVEL    GEARS.  393 

Measuring  the  back  cone  radius  of  the  small  gear,  it  is 
found  to  be  f  ^  inch,  and  the  diameter  will  be  f  |  inch ;  conse- 
quently the  shape  of  tooth  at  a  for  the  small  gear  is  the  same  as 
the  shape  of  tooth  in  a  spur  gear  of  21  teeth,  sixteen  diametral 
pitch. 

Therefore,  if  this  pair  of  gears  is  to  be  cut  by  a  rotary 
cutter  having  a  fixed  curve,  a  different  cutter  is  required  for 
each  gear. 

When,  in  a  pair  of  bevel  gears,  both  gears  are  of  the  same 
size  and  have  the  same  number  of  teeth,  and  their  axial  center 
lines  are  at  right  angles,  they  are  called  miter  gears,  and  one 
cutter,  of  course,  will  answer  for  both  gears.  One  cutter  will 
also  answer  in  practice  when  the  difference  of  the  back  cone 
radius  of  a  pair  of  gears  is  so  small  that  it  comes  within  the 
limit  of  one  cutter  as  used  for  spur  gears  of  the  same  size.  Bevel 
gears  may  also  be  made  with  cycloid  form  of  teeth,  but  when- 
ever cut  by  rotary  cutters,  as  usually  employed  in  producing 
small  bevel  gears  of  diametral  pitch,  the  involute  form  of  tooth 
should  always  be  used. 

Cutting  Bevel  Gears. 

When  bevel  gear  teeth  are  correctly  formed,  the  tooth 
curve  will  constantly  change,  from  one  end  of  the  tooth  to  the 
other.  Therefore,  bevel  gears  of  theoretically  correct  form 
cannot  be  produced  by  a  cutter  of  fixed  curve ;  but,  practically, 
very  satisfactory  results  are  obtained  in  cutting  bevel  gears  of 
small  and  medium  size  in  this  way. 

When  a  regular  gear-cutting  machine  is  not  at  hand,  the 
Universal  milling  machine  is  a  very  convenient  tool  for  cutting 
bevel  gears  of  moderate  size,  and  is  used  in  the  following  way : 

First,  see  that  the  gear  blank  is  turned  to  correct  size  and 
angle,  and  adjust  the  machine  to  the  angle  corresponding  to  the 
bottom  of  the  teeth  in  the  gear.  The  correct  index  is  set  ac- 
cording to  the  number  of  teeth  in  the  gear.  Adjust  the  cutter 
to  come  right  to  the  centre  of  the  gear,  cut  the  correct  depth  as 
marked  on  the  gear  at  a  (see  Fig.  9),  according  to  Table  No.  49, 
and  when  the  machine  is  adjusted  to  the  correct  angle,  and  the 
correct  depth  is  cut  at  a,  the  correct  depth  at  b  will,  as  a  matter 
of  fact,  be  obtained. 

Second,  when  a  few  teeth  are  cut  in  the  gear  (two  or  three) 
bring,  by  means  of  the  index,  the  first  tooth  back  to  the  cutter. 
By  means  ©f  the  index,  rotate  the  gear,  moving  the  tooth  toward 
the  cutter ;  but,  by  the  slide,  move  the  gear  sidewise  away  from 
the  cutter,  until  the  cutter  coincides  with  the  space'at  b ;  then  cut 
through  from  a  to  b.  This  operation  will  widen  one  side  of  the 
tooth  space  at  a. 

Note  the  position  of  the  machine,  and,  by  the  use  of  the 
index  and  slide,  return  the  cutter  to  its  central  position  and  in- 


394  BEVEL    GEARS. 

dex  into  the  next  space,  and  rotate  the  other  side  of  the  tooth 
toward  the  cutter  as  much  as  the  first  side ;  but,  by  the  slide, 
the  gear  is  moved  sidewise  away  from  the  cutter  until  the 
cutter  coincides  with  the  space  at  b;  then  cut  through  on  this  side 
from  a  to  b.  Thus,  by  repeated  cutting  on  each  side  alternate- 
ly, one  tooth  is  backed  off  equally  on  both  sides  and  measured 
by  a  gage,  until  the  correct  thickness  on  the  pitch-line  at  a, 
according  to  Table  No.  49,  is  obtained. 

Be  very  careful  to  have  the  machine  set  over  the  same 
amount  on  each  side  of  the  tooth,  or  else  the  tooth  will  be 
askew. 

Third,  when  one  tooth,  thus  by  trial,  is  correctly  cut, 
note  the  position  of  the  machine  and  cut  all  the  teeth  through 
on  one  side,  then  set  over  to  the  other  side  in  exactly  the 
same  position  as  was  found  to  be  right  for  the  first  tooth  ;  cut 
through  again  and  the  gear  is  finished.  Thus,  when  the  correct 
position  of  the  machine  is  obtained,  any  number  of  gears  of 
the  same  size  and  same  pitch  may  be  cut,  by  simply  letting  the 
cutter  go  through  twice. 

NOTE. — As  already  stated,  bevel  gear  cutting  in  this  way  is 
only  a  compromise  at  the  best,  but  by  careful  manipulation 
and  good  judgment  an  experienced  man  is  able  to  do  a  very 
creditable  job.  A  cutter  is  usually  selected  of  the  same  curve 
as  is  correct  for  a  spur  gear  corresponding  to  the  back  cone 
radius  of  the  gear.  Thus,  it  may  be  thought  that  the  shape 
of  the  tooth  should  be  the  shape  of  the  cutter,  but  by  investi- 
gation it  will  be  found  that,  on  account  of  the  "  backing  off," 
the  teeth  will  be  of  a  little  more  rounding  shape  at  the  large  end 
than  corresponds  to  the  cutter;  therefore,  when  the  gear  has  few 
teeth, — less  than  25, — it  is  usually  preferable  to  make  the  shape 
of  the  cutter  to  correspond  to  a  gear  a  little  larger  than  would 
be  called  for  by  the  back  cone  radius  of  the  bevel  gear  to  be 
cut;  but  when  the  gear  has  more  than  25  teeth,  a  cutter  of  shape 
corresponding  to  the  back  cone  radius  of  the  gear  will  give 

fDod  results.  For  instance,  in  the  pair  of  bevel  gears  shown  in 
ig.  9  the  back  cone  radius  of  the  large  gear  calls  for  a 
cutter  corresponding  in  shape  to  a  cutter  for  a  spur  gear  of  59 
teeth,  16  diametral  pitch ;  and  this  shape  of  cutter  will,  after 
the  teeth  are  backed  off,  make  the  teeth  a  trifle  too  round  at 
the  large  end,  and  a  trifle  too  straight  on  the  small  end,  but  if 
the  teeth  are  not  too  long  the  job  will  be  very  satisfactory. 

The  back  cone  radius  of  the  small  gear  calls  for  a  cutter 
corresponding  in  shape  to  a  cutter  for  a  spur  gear  of  21  teeth, 
16  diametral  pitch,  but  when  the  teeth  are  backed  off  they 
will  be  a  little  too  rounding  on  the  large  end  ;  therefore  a  better 
result  is  obtained  by  selecting  a  cutter  having  a  shape  corre- 
sponding to  a  little  larger  spur  gear ;  for  instance,  a  gear  of 
24  teeth.  Such  a  cutter  will  give  the  teeth  a  better  shape  on 


BEVEL    GEARS. 


395 


the  large  end,  although  it  may  be  necessary  to  round  the  teeth 
a  little,  outside  the  pitch  line  on  the  small  end,  by  filing. 

Of  course,  a  spur  gear  cutter  cannot  be  used  for  cut- 
ting bevel  gears,  because,  although  it  may  have  the  correct 
curve,  it  would  be  too  thick.  The  thickness  of  a  bevel  gear 
cutter  must  be  at  least  0.005  inch  thinner  than  the  space  be- 
tween the  teeth  at  their  small  end. 

Large  bevel  gears  are  made  on  theoretically  correct  prin- 
ciples by  planing  on  specially  constructed  machines. 


WORMS  AND  WORM  GEARS. 

Fig.  10  shows  a  worm  and  worm  gear. 


7TF 


f=  Pitch  diameter  of  gear. 

f  =  Smallest  outside  diameter. 
=  Largest  outside  diameter. 
a  =  Outside  diameter  of  worm. 
b  =  Pitch  diameter  of  worm. 
c  =  Diameter  of  worm  at  bottom  of  thread. 

The  pitch  and  diameter  of  worm  screws  are  usually  of  such 
proportions  that  for  single-thread  the  angle  of  the  teeth  on  the 
gear  is  from  two  to  three  degrees.  This  angle  is  most  conven- 
iently obtained  by  drawing  a  diagram  as  shown  in  Fig.  10. 


WORMS    AND    WORM    GEARS. 


Draw  a  line  ////,  equal  to  31  times  the  length  of  line  b',  this  line 
will  be  equal  to  the  length  of  the  circumference  of  the  pitch 
diameter  of  the  screw.  Erect  the  perpendicular,  m  o,  equal  to 
the  pitch  of  the  screw.  Connect  the  points  /and  o  by  the  line 
/  <?,  and  the  angle  s  is  the  angle  of  the  teeth  on  the  worm  gear. 

NOTE. — When  the  pitch  diameter  of  the  worm  screw  is 
seven  times  the  circular  pitch  of  the  worm  gear  and  the  worm 
has  single  thread,  the  angle  of  the  thread  on  the  gear  is  very 
nearly  2>^  degrees. 

CAUTION. — When  cutting  a  worm  gear,  be  careful  and  not 
lay  the  angle  of  the  teeth  in  the  wrong'direction. 

The  diameter  of  worm  gears  is  usually  calculated  according 
to  circular  pitch,  for  convenience  in  cutting  the  worm  with  the 
same  gears  as  used  for  ordinary  screw  cutting  in  a  lathe. 

When  a  worm  gear  has  comparatively  few  teeth,  the  flank 
of  the  tooth  will  be  undercut  by  the  hob;  to  prevent  this  in  a 
measure,  it  is  customary  to  have  the  blank  somewhat  over  size, 
so  that  from  five-eighths  to  three-fourths  of  the  depth  of  the  tooth 
may  be  outside  the  pitch  line. 

The  form  of  teeth  is  usually  involute,  and  the  thread 
on  a  worm  screw  is  constructed  of  the  same  shape  as  the  teeth 
in  a  rack.  Fig.  11  shows  the  shape  of  tooth  and  the  tabh 
gives  the  dimensions  of  finishing  tool  for  the  most  common 
pitches. 

The  surface  speed  of  a  worm  screw  ought  not  to  exceed  300 
feet  per  minute. 

Table  No.  50  is  calculated  by  the  following  formulas.  (See 
Fig.  11.) 

P  =  Circular  pitch. 
A'=  — 

"  p 


a  =  P  X  0.3183 
d  —  P  X  0.3683 


S  =  P  X  0.5 
b  =  P  X  0.31 
C=P  X  0.335 


k  =  P  X  0.1 


WORMS    AND    WORM    GEARS. 


397 


TABLE  No.  50.— Giving  Proportions  of  Parts  for  Worms 
and  Worm  Gears,  Calculated  According  to  Circular  Pitch. 

(See  Fig.  11.) 


•UUOM.  JO 

ip  J3AO  qoq  jo 

J3)3UIBIp    til 


_  >O  O  »O         iO  CO  CO  GO  »»  -M  O  t~  CO 

O  1~  »O  CN         l-SD»O-*CO''N<M<N<MT-iTHi-t 
fN-HrH-^rHOOOOOOOOOOOO 


S^COOOOCOrHOSGOl-CpiO^CO 
O-fCO^^i-^— 'OOOOOOO 


JO  t{»d3Q  Sui^JO^V 


•innpnappy 


-BIQ  Bui 


•qoui  aad 


•saqout 


^O         kO  OS  CO  1^*  OS  ^1 
«O  t-  t-  1--  GOGOOCOS 

53  S  t-J  TH 


t  GO  1^  SO  »O  CO  CO  C<l 


(M  iH  iH  TH  T-l 


393 


WORMS    AXD    WORM    GEARS. 


TABLE  No.  51.— Showing  How  to  Gear  Lathes  when  Cut= 
ting  Worms  of  the  Pitches  Given  in  Table  No.  50. 


2 

la 

sl 

?  OF  THREADS 
*  INCH. 

LEADING  SCREW  OF   LATHE. 

a 

Threads 
per  inch. 

3 

Threads 
per  inch. 

4 

Threads 
per  inch. 

5 

Threads 
per  inch. 

6 

Threads 
per  inch. 

10 

Threads 
per  inch. 

o  . 

w  w 

t 

u 

c3 

• 

* 

•§! 

|| 

•a  8 

3  bfl 

s| 

it 

\i 

"a! 

\i 

•§! 

II 

"sl 

ii 

o> 

in 

C/3 

C/3 

^ 

c/5 

& 

CO 

CO 

c^ 

in 

2 

% 

160 

40 

1  *// 

^7 

140 

40 

1  V^ 

2/ 

120 

40 

ji5 

% 

100 

40 

120 

32 

1 

1 

80 

40 

96 

32 

1/^3 

60 

40 

72 

32 

120 

40 

% 

1% 

50 

40 

60 

32 

100 

40 

i/ 

2 

40 

40 

.48 

32 

80 

40 

100 

40 

% 

2% 

40 

50 

48 

40 

64 

40 

80 

40 

i/ 

3 

40 

60 

48 

48 

64 

48 

60 

36 

48 

24 

2/7 

g^/ 

40 

70 

48 

56 

64 

56 

60 

42 

48 

28 

4 

40 

80 

24 

32 

40 

40 

50 

40 

48 

32 

% 

4i/ 

40 

90 

24 

36 

32 

36 

40 

36 

48 

36 

M> 

5 

20 

50 

24 

40 

32 

40  j 

40 

40 

48 

40 

60 

30 

% 

6 

20 

60 

24 

48 

32 

48  1 

40 

48 

48 

48 

60 

36 

8 

20 

80 

24 

64 

24 

48 

30 

48 

48 

64  JCO 

48 

rV 

10 

'    20 

100 

24 

80 

24 

60 

24 

48 

24 

40||50 

50 

Reduction  of  Speed  by  Worm  Gearing. 

In  a  single-threaded  worm  screw  one  revolution  of  the  worm 
moves  the  gear  one  tooth ;  in  a  double-threaded  worm  screw  one 
revolution  of  the  worm  moves  the  gear  two  teeth,  and  in  a  triple- 
threaded  worm  screw  one  revolution  of  the  worm  moves  the 
gear  three  teeth.  A  great  deal  ol  work  is  lost  by  friction  by 
using  worm  gearing,  frequently  from  50  to  75  per  cent.,  some  of 
which  could  be  saved  by  using  a  ball  bearing  to  take  the  end 
thrust  of  the  worm.  The  efficiency  is  also  increased  by  using  a 
worm  of  double,  triple  or  quadruple  thread,  because  this  in- 
creases the  angle  of  the  teeth  in  the  wheel  and  the  efficiency  of 
the  mechanism  is  increased  by  increasing  the  angle  until  it 
reaches  20°  to  25°,  when  it  rapidly  falls  off  again. 

Calculating  the  Size  of  Worm  Gears. 

EXAMPLE. 

Find  dimensions  of  a  worm  gear  having  68  teeth,  ^  inch 
pitch,  cut  teeth.  Make  the  pitch  diameter  of  the  single-thread 
worm  six  times  the  pitch  of  the  worm,  Use  Table  No,  50, 


WORMS   AND    WORM   GEARS.  399 

Solution  : 

In  Table  No.  47  the  pitch  diameter  of  a  gear  of  68  teeth  of 
one-inch  pitch  is  given  as  21.65  inches;  thus,  the  pitch  diameter 
for  a  gear  of  68  teeth  of  |<-inch  circular  pitch  will  be  21.65  X 
0.75  =  16.788  inches. 

In  column  a,  of  Table  No.  50,  the  addendum  for  %-inch 
circular  pitch  is  given  as  0.2387  inch  ;  fhis  is  multiplied  by  2, 
because  it  is  to  be  added  on  both  sides  of  the  gear. 

Thus,  the  smallest  outside  diameter  of  the  gear  is  16.738  + 
0.2387  X  2  =  17.215  inches;  or,  practically,  17^  inches.  If  the 
gear  is  to  be  made  hollow  to  correspond  to  the  "curve  at  bottom 
of  thread  of  the  worm,  make  a  scale  drawing  as  shown  in  Fig. 
10,  and  make  line^-,  17^  inches  ;  from  this  drawing  the  largest 
outside  diameter  may  be  obtained  by  measurement. 

The  diameter  of  the  worm  on  the  pitch  line  was  to  be  six 
times  the  pitch  =  6  X  %"  =  4^  inches. 

The  addendum  for  the  thread  on  the  worm  can  be  obtained 
from  Table  No.  50,  column  *z,  and  is  0.2387.  The  outside 
diameter  of  the  worm  will  be  4.5  -f  2  X  0.2387  =  4.977  inches,  or, 
practically,  4f£  inches. 

The  cutter  to  be  used  in  roughing  out  the  gear  should  have 
a  curve  of  involute  form  corresponding  to  a  spur  gear  cutter  for 
68  teeth,  and  its  thickness  ought  to  be  at  least  0.005  inch  less 
than  the  width  of  space  as  given  in  column  S  of  Table  No.  50. 
Therefore  the  thickness  on  the  pitch  line  of  the  roughing  cutter 
will  be  0.37  inch. 

The  angle  of  the  teeth  may  be  obtained  from  a  drawing  as 
shown  and  explained  in  Fig.  10,  or  it  may  be  calculated  thus  : 

circular  pitch 
Tangent  of  angle  S  =  —  —, 

pitch  circumference 


In  Table  No.  21  the  corresponding  angle  is  given  as  3 
degrees,  very  nearly. 

The  depth  to  which  the  gear  should  be  cut  is  given  in  col- 
umn D  as  0.515  inch.  The  gear  is  finished  with  a  hob,  as 
described  below,  which  is  allowed  to  cut  until  it  touches  the 
bottom  of  the  spaces  in  the  gear.  The  outside  diameter  of  the 
hob  should  be  larger  than  the  outside  diameter  of  the  worm,  in 
order  that  the  teeth  in  the  hob  may  reach  the  bottom  of  the 
spaces  in  the  gear  and  leave  clearance  for  the  worm,  and  at  the 
same  time  leave  the  gear  tooth  of  the  proper  thickness  on  the 
pitch  line.  This  increment  is  obtained  in  column  £,  Table  No. 
50,  and  for  %-inch  pitch  is  0.075  inch  ;  thus,  the  outside  diameter 
of  the  hob  is  0.075  inch  larger  than  the  outside  diameter  of  the 
worm,  or  4.977  +  0.075  =  5.052  inches.  The  angle  of  the  finish- 
ing threading  tool  for  both  worm  and  hob  is  14)4  degrees, 
making  the  angle  of  space  29°,  as  shown  in  Fig.  11.  The  clear- 
ance angle  of  the  threading  tool  must  be  a  little  more  than  the 
angle  of  the  thread. 


4oo 


WORMS   AND   WORM    GEARS. 


The  width  of  the  threading  tool  at  the  point  is  given  in 
Column  b,  Table  No.  50,  as  0.2325  inch.    The  depth  of  the  space 
to  be  cut  in  the  worm  is  given  in  Column  /?,  as  0.515  inch.     The 
diameter  of  the  worm  at  the  bottom  of  the  thread  will  be : 
4.977  —  2  X  0.515  =  3.947  inches. 

The  depth  of  the  space  to  be  cut  in  the  hob  is  given  in  col- 
umn h  in  Table  No.  50  as  0.5525  inch. 

The  diameter  of  the  hob  at  bottom  of  thread  will  be : 
5.052  —  2  X  0.5525  =  3.947  inches. 

Thus  the  only  difference  in  size  between  the  hob  and  the 
worm  is  in  the  outside  diameter  and  in  the  depth  of  the  cut. 
Both  may  be  finished  by  the  same  tool,  as  the  diameter  at  the 
bottom  of  the  thread  and  the  thickness  of  the  teeth  at  the  pitch 
line  should  be  the  same  for  both  hob  and  worm. 

Elliptical  Gear  Wheels. 

Elliptical  gear  wheels  are  sometimes  used  in  order  to  change 
a  uniform  rotary  motion  of  one  shaft  to  an  alternately  fast  and 
show  motion  of  the  other.  See  Fig.  12. 


The  pitch  line  is  constructed  and  calculated  the  same  as 
the  circumference  of  an  ellipse.  (See  page  189.)  The  gear  is 
constructed  involute  the  same  as  for  spur  gears.  If  the  differ- 
ence between  the  minor  and  the  major  diameters  is  large  it  may 
be  necessary  to  construct  the  teeth  of  different  shapes  at  differ- 
ent places  on  the  circumference;  in  other  words,  the  whole  cir- 
cumference of  the  gear  cannot  be  cut  with  the  same  cutter.  A 
cutter  of  the  same  pitch,  of  course,  but  corresponding  to  a  larger 
diameter  of  gear,  must  be  used  where  the  curve  of  the  pitch  line 
is  less  sharp. 

The  centers  of  the  shafts  are  in  the  foci  of  the  ellipse.  If 
two  elliptical  gear  wheels,  made  from  the  same  pattern,  or  cut 
together  at  the  same  time,  on  the  same  arbor,  are  to  work  together 
they  must  have  an  uneven  number  of  teeth  so  that  a  space  will 
be  diametrically  opposite  a  tooth,  as  will  be  seen  from  Fig.  12, 


SCREWS.  4OI 

SCREWS. 

"Pitch,"  "Inch  Pitch"  and  "Lead"  of  Screws  and  Worms. 

The  term  "  pitch  of  a  screw,"  as  commonly  used,  means  its 
number  of  threads  per  inch,  while  the  "  inch  pitch  "  is  the  dis- 
tance from  the  center  of  one  thread  to  that  of  the  next.  For 
instance,  a  one-inch  screw  of  standard  thread  is  usually  said  to 
be  an  "  eight  pitch"  screw,  because  it  has  eight  threads  per  inch  of 
length ;  but  it  might  more  correctly  be  said  to  be  a  screw  of 
>£-inch  pitch,  because  it  is  >£-inch  from  the  center  of  one  thread 
to  the  center  of  the  next. 

The  "  lead  "  of  a  worm  or  a  screw  means  the  advancement 
of  the  thread  in  one  complete  revolution ;  therefore,  in  a  single- 
threaded  screw,  the  inch  pitch  and  the  lead  is  the  same  thing,  but 
in  a  double  or  triple-threaded  screw  the  inch  pitch  and  the  lead  are 
two  different  things.  The  "  lead  "  in  a  double-threaded  screw  will 
be  a  distance  equal  to  twice  the  distance  from  the  center  of  one 
thread  to  the  center  of  the  next,  but  in  a  triple-threaded  screw  the 
lead  is  three  times  the  distance  from  the  center  of  one  thread  to 
the  center  of  the  next. 

Screw  Cutting  by  the  Engine  Lathe. 

When  the  stud  and  the  spindle  run  at  the  same  speed 
(  which  they  usually  do)  the  ratio  between  the  gears  may  always 
be  obtained  by  simply  ascertaining  the  ratio  between  the  num- 
ber of  threads  per  inch  of  the  lead-screw  and  the  screw  to  be 
cut. 

EXAMPLE. 

The  lead-screw  on  a  lathe  has  four  threads  per  inch  and 
the  screw  to  be  cut  has  \ll/2  threads  per  inch  ( one-inch  pipe- 
thread).  Find  the  gears  to  be  used  when  the  smallest  change  gear 
lias  24  teeth  and  the  gears  advance  by  four  teeth  up  to  96. 
The  ratio  of  the  number  of  threads  per  inch  of  the  two  screws 
is  as  4  to  \\yz. 

As  the  smallest  gear  has  24  teeth  and  the  gears  all 
advance  by  four  teeth,  this  ratio  of  the  screws  must  be  mul- 
tiplied by  a  number  which  is  a  multiple  of  4  and  which,  at  least, 
gives  the  smallest  gear  24  teeth.  For  instance,  multiply  by  8 
and  the  result  is  8  X  11  ^  =  92  teeth  for  the  gear  on  the  lead- 
screw  ;  8  X  4  =  32  teeth  for  the  gear  on  the  stud. 

Cutting  flultipIe-Threaded  Screws  or  Nuts  by  the 
Engine  Lathe. 

Calculate  the  change  gears  as  if  it  was  a  single-threaded 
screw  of  the  same  lead.  Cut  one  thread  and  move  the  tool  the 
proper  distance  and  cut  the  next  thread. 


4O2  SCREWS. 

The  most  practical  way  to  move  the  tool  from  one  thread 
to  another,  when  cutting  double-threaded  screws  or  nuts,  is 
to  select  a  gear  for  the  stud  or  spindle  of  the  lathe  having  a 
number  of  teeth  which  is  divisible  by  two,  and  when  one  thread  is 
cut  make  a  chalk  mark  across  a  tooth  in  this  gear  onto  the  rim  of 
the  intermediate  gear  ;  count  half  way  around  the  gear  on  the  stud 
and  make  a  chalk  mark  across  that  tooth;  drop  the  swing  plate 
enough  to  separate  the  gears,  pull  the  belt  by  hand  until  the  oppo- 
site mark  on  the  gear  on  the  stud  comes  in  position  to  match  the 
chalk  mark  on  the  intermediate  gear;  clamp  the  swing  plate  again 
and  the  tool  is  in  proper  position  to  cut  the  second  thread. 

When  triple  threads  are  to  be  cut,  select  a  gear  for  the 
spindle  or  stud  whose  number  of  teeth  is  divisible  by  three, 
and  in  changing  the  tool  from  one  thread  to  the  next,  only  turn 
the  lathe  enough  so  that  the  gear  on  the  stud  moves  one-third 
of  one  revolution. 

If,  for  any  reason,  it  should  be  inconvenient  to  make  this 
change  by  the  gear  on  the  stud,  the  change  may  be  made  by  the 
lead-screw  gear.  The  intermediate  gear  is  first  released  from  the 
gear  on  the  lead-screw,  which  is  then  moved  ahead  the  proper 
number  of  teeth,  and  again  connected  with  the  intermediate  gear. 
The  proper  number  of  teeth  to  move  the  gear  on  the  lead-screw 
is  obtained  by  the  following  rule  : 

Multiply  the  number  of  teeth  in  the  gear  on  the  lead-screw 
by  the  number  of  threads  per  inch  of  the  lead-screw  ;  divide  this 
product  by  the  number  of  threads  per  inch  of  the  screw  to  be 
cut,  and  the  quotient  is  the  number  of  teeth  that  the  gear  on  the 
lead-screw  must  be  moved  ahead. 

EXAMPLE. 

A  square-threaded  screw  is  to  have  ^-inch  lead  and  triple 
thread.  The  lead-screw  in  the  lathe  has  two  threads  per  inch, 
and  the  gear  on  the  lead  screw  has  ninety-six  teeth.  How  many 
teeth  must  the  gear  on  the  lead-screw  be  moved,  when  changing 
from  one  thread  to  the  next  ? 

Solution  : 

A  screw  of  J^-inch  lead  with  triple  thread  has  six  threads 

2  X  96 
per  inch,  therefore  the  gear  must  be  moved  —  ^  —  =  32  teeth, 

in  order  to  change  the  tool  from  one  thread  to  the  next. 


U.  S.  Standard  Screws. 

Fig.  1  shows  the  shape  of 
thread  on  United  States  stand- 
ard screws.  The  sides  are 
straight  and  form  an  angle  of 
sixty  degrees,  and  the  thread  is 


SCREWS.  403 

flat  at  the  top  and  bottom  for  a  distance  equal  to  one-eighth  of 
the  pitch,  thus  the  depth  of  the  thread  is  only  three-fourths  of 
a  full,  sharp  thread.  (See  Fig.  1.) 

Fig.  2  shows  the  shape  of  the  Whitworth  (the  English)  sys- 
tem of  thread.  As  compared 
with  the  American  system, 
the  principal  difference  is  in 
the  angle  between  the  sides 
of  the  thread,  which  is  fifty- 
five  degrees,  and  one-sixth  of 
the  depth  of  the  full,  sharp 
thread  is  made  rounding  at 
the  top  and  bottom.  There  is  also  a  difference  in  the  pitch  of 
a  few  sizes. 

The  common  V-thread  screws  have  the  angle  of  thread  of 
sixty  degrees,  the  same  as  the  United  States  standard  screws, 
but  the  thread  is  sharp  at  both  top  and  bottom.  This  style  of 
thread  is  rapidly,  as  it  should  be,  going  out  of  use.  The  prin- 
cipal disadvantages  of  this  thread  are  that  the  screw  has  less 
tensile  strength,  and  it  is  also  very  difficult  to  keep  a  sharp- 
pointed  threading  tool  in  order. 


Diameter  of  Screw  at  Bottom  of  Thread. 

The  diameter  of  screws  at  the  bottom  of  thread  is  obtained 
by  the  following  formulas  :  — 

United  States  Standard  Screws: 


n 
For  V-threaded  screws  : 


n 
For  Whitworth  screws  : 


n 

d=  Diameter  of  screw  at  bottom  of  thread. 
D  =  Outside  diameter  of  screw. 
n  =  Number  of  threads  per  inch. 
1.299  is  constant  for  United  States  standard  thread. 
1.733  is  constant  for  sharp  V-thread. 
1.281  is  constant  for  Whitworth  thread. 


404 


SCREWS. 


TABLE  No.  52.— Dimensions  of  Whitworth  Screws. 


Diameter  of 
Screw  in 
Inches. 


H 


Number  of 

Threads  per 

Inch. 


40 

24 

20 

18 

16 

14 

12 

11 

10 

9 

8 

7 


Diameter   of    Number  of 
Screw  in     Threads    per 
Inches.  Lich. 


Diameter  of 
Screw  in 
Inches. 


3% 
4 


4% 
5 


Number  of 
Threads  per 
Inch. 


Diameter  of  Tap  Drill. 

The  diameter  of  the  drill  with  which  to  drill  for  a  tap  is,  if 
we  want  full  thread  in  the  nut,  equal  to  the  diameter  of  the 
screw  at  the  bottom  of  the  thread,  and  is,  therefore,  obtained  by 
the  same  formulas.  However,  in  practical  work  it  is  always  ad- 
visable to  use  a  drill  a  little  larger  than  the  diameter  of  the 
screw  at  the  bottom  of  thread,  because  in  threading  wrought  iron 
or  steel  the  thread  will  swell  out  more  or  less,  and  a  few  thou- 
sandths must  be  allowed  in  the  size  when  drilling  the  hole.  In 
drilling  holes  for  tapping  cast-iron,  a  little  larger  drill  is  used, 
because  it  is  unnecessary  in  a  cast-iron  nut  to  have  exactly  full 
thread.  Table  No.  53  gives  sizes  of  drill  for  both  wrought  and 
cast-iron,  which  give  good  practical  results  for  United  States 
standard  screws. 

Table  No.  53  gives  sizes  of  hexagon  bolts  and  nuts.  The 
size  of  the  hexagon  is  equal  to  1)4  times  the  diameter  of  bolt  + 
^-inch ;  the  thickness  of  head  is  equal  to  half  the  hexagon. 
The  thickness  of  nut  is  equal  to  the  diameter  of  the  bolt. 
When  heads  and  nuts  are  finished  they  are  ^-inch  smaller. 

The  table  is  calculated  by  the  following  formulas : 


d  =  D 


1.299 


C  =  1.155  A 


B  =  1AUA 
F=D 


SCREWS.  405 

TABLE  No.  53. — Dimensions  of  U.  S.  Standard  Screws. 


X 


u  « 

s* 


0.185 
0.240 
0.294 
0.345 
0.400 
0.454 
0.507 
0.620 
#  0.731 
0.838 
0.939 
1.065 
1.158 
1.284 
1.389, 
1.490 
1.615 
1.711 
1.961 


4tf 

i^2 

'-V 

5# 


2.175  2A 
2.425  2X 
2.629  2  d 
2.8792§  j 
3.1003i^ 
3.3173 
3.567 
3.798  3j 
4.028  4£ 
4.255  4-,»> 
4.480  4^ 
4.730'4^ 
4.963  4f| 
5.2035|| 
5.4236*1 


0.0269 

0.0452 

0.0679 

0.0935 

0.1257 

0-1619 

0.2019 

0.3019 

0.4197 

0.5515 

0.6925 

0.8892 

1.0532 

1.2928 

1.5153 

1.7437 

2.0485 

2.2993 

3.0203 

3.7154 

4.6186 

5.4284 

6.5099 

7.5477 

8.6414 

9.9930 

11.3292 

12.7366 

14.2197 

15.7633 

17.5717 

19.2676 

21.2(U7 

23.0978 


0.0062 
0.0069 
0.0078 
0.0089 
0.0096 
0.0104 
0.0114 
0.0125 
0.0139 
0.0156 
0.0178 
0.0178 
0.0208 
0.0208 
0.0227 
0.0250 
0.0250 
0.0278 
0.0278 
0.0313 
0.0313 
0.0357 
0.0357 
0.0384 
0.0417 
0.0417 
0.0435 
0.0455 
0.0476 
0.0500 
0.0500 
0.0526 
0.0526 
0.0556 


406 


SCREWS. 


NOTE. — In  finished  work,  the  thickness  of  the  head  of  the 
bolt  and  the  nut  is  equal,  and  is  Vie  of  an  inch  less  than  the 
diameter  of  the  bolt. 

Columns  B  and  C  in  Table  No.  53  are  very  useful  for  many 
purposes ;  for  instance,  in  selecting  size  of  counter-bore  when 
finishing  castings,  to  give  bearing  for  screw  heads  :  in  turning 
blanks  which  are  afterwards  to  be  cut  into  square  or  hexagon 
heads,  etc. 

Table  No.  54.— Coupling  Bolts  and  Nuts. 

(Hexagon). 
(All  Dimensions  in  Inches) 


kl 

<u 

«d 

Dimensions  of  Head  and  Nut. 

1) 

rt   HH 

1 

S 

1* 

Across 
the  Flat. 

Across 
the 
Corner. 

Length 
of  Head 
or  Nut. 

y?, 

13 

^ 

IJL 

y* 

5/8 

11 

lyV 

HI 

5/8 

24 

10 

li^ 

iTv 

li 

^ 

9 

IA 

% 

i 

8 

15/8 

i^ 

1 

1^6 

7 

Hi 

2/2 

li/6 

ITA 

7 

2 

2T% 

!^ 

Table  No.  55.— Round  and  Fillister  Head  Screws, 

(All  Dimensions  in  Inches). 


Diameter  of 
Screw. 

Number  of 
Threads  per 
Inch. 

Diameter  of 
Head. 

Length  of 
Head. 

^ 

40 

A 

H 

T3?r 

24 

i^ 

T3F 

i/ 

20 

3A 

I/ 

tV 

18 

7 

A 

3^ 

16 

J^. 

3/^ 

A 

14 

5/4 

13 

M 

i^ 

T%" 

12 

If 

y9_. 

5/^ 

11 

% 

5/6 

24 

10 

1 

M 

^ 

9 

1^ 

^ 

i 

8 

^ 

1                ] 

J 

Round   Head 
Cap  Screw. 


Fillister  Head 


SCREWS. 


407 


TABLE  No.  56.— Dimensions  of  Hexagon  and  Square  Head 
Cap  Screws. 

(All  Dimensions  in  Inches). 


'o 

cL 

Hexagon  Head. 

Square  Head. 

o 

<L>    > 

en 

_: 

QJ     Q) 

'"C  <-< 

"£  rt 

s  b 

2| 

Across 

Across 

Across 

Across 

bflOJ 

.2  ^2 

the 

the 

the 

the 

8* 

Q 

P 

Flats. 

Corners. 

Flats. 

Corners. 

1-1 

X 

20 

A 

y, 

3/8 

Ji 

5< 

A 

18 

IJ 

TV 

5/^ 

T*V 

3/8 

16 

A 

» 

/^ 

tl 

3/8 

A 

14 

5/8 

II 

A 

«i 

TV 

13 

y\ 

II 

5/8 

li 

y2 

A 

12 

H 

H 

IJ 

A 

H 

11 

IJlj 

j| 

ly1^ 

M* 

10 

i 

IA 

^ 

ll^ 

^ 

4 

9 

1/8 

HI 

1/8 

HI 

^ 

i 

8 

1* 

1/i 

i 

1/8 

7 

13/8 

HI 

13/8 

HI 

ix/6 

*# 

7 

l1^ 

Hi 

1^ 

%l/8 

i/< 

Eye  'Bolts. 

It  is  very  customary  to  weld  an  eye  to  a  lag  screw  (see  Fig. 
3)  to  use  in  handling  heavy  weights  in  shops. 

The  following  table  (No.  57)  gives  the  holding  power  of  lag 
screws  or  eye  bolts  when  screwed  into  spruce  timber  a  little 
over  the  full  length  of  thread.  The  suitable  size  of  bit  for  the 
thread  is  also  given  in  the  table. 

TABLE  No.  57- 


Diameter 
of  Screw. 

Diameter 
of  Bit. 

Load  at 
Which  the 
Screw 
Pulled  Out. 

Safe 
Load. 

1  inch 

y±  inch 

16,000  Ibs. 

2,000  Ibs 

7/8     " 

«    " 

9,000    " 

1,125  " 

34     « 

5/8     " 

7,000    " 

875  " 

5/8     " 

X   " 

6,000    « 

750  " 

#     " 

3/8      " 

3,500    " 

437  " 

3/8     " 

-&      " 

1,900    « 

237  « 

X  " 

&      " 

700    « 

87  " 

408 


SCREWS. 


TABLE  No.  58.— Giving  the  Average  Weight  in  Pouncj 
per  100  Square  Head  ( iim let = Pointed  Lag  Screws. 


LENGTH 

in  Inches. 

ir 

A" 

3/8" 

7    " 

#" 

ft" 

«" 

y." 

X* 

1" 

15* 

254 

H 

7 

10 

2 

3^2 

H 

8i 

12 

17 

24 

27^ 

2^ 

4j^ 

S* 

9| 

14 

19 

26 

31 

3 

43^ 

11 

16 

21 

28 

34 

51 

zy* 

5k 

8* 

121 

18 

24 

31 

38 

55 

A. 

53/£ 

9* 

14 

20 

26 

34 

42 

60 

85 

112 

4*^ 

6^ 

10i 

151 

22 

28 

37 

46 

65 

91 

121 

5 

7 

Hi 

17 

24 

32 

40 

50 

70 

97 

130 

5% 

~*% 

12t 

18* 

26 

34 

43 

54 

76 

103 

140 

6 

8 

20 

28 

36 

46 

58 

81 

110 

150 

6^ 

21^ 

30 

38 

49 

62 

86 

117 

100 

7 

23 

32 

41 

52 

65 

92 

125 

170 

7¥ 

241 

34 

44 

55 

69 

97 

132 

iso 

8 

26 

36 

47 

58 

73 

103 

140 

190 

&l/2 

77 

108 

148 

lOO 

9 

81 

113 

156 

2ir 

9/^j 

• 

85 

118 

164 

22' 

10 

89 

123 

172 

23d 

Size  of 

* 

^ 

•6 

33 

* 

<> 

N, 

SJH 

l/ts 

^ 

RfcO 

•-r-1 

Head  in 

X 

X 

X 

X 

X 

X 

X 

X 

X 

Inches. 

£ 

X 

MM 

«|w 

T—  1 

WH 

* 

GIMLET-POINTED    LAG    SCREW. 


EXAMPLE. 

What  is  the  weight    of  8  lag    screws  6"  long  and 
in  diameter  ? 

Solution: 

Under  the  heading  ]/2  -inch,  in  the  line  with  6  in  the  column 
of  length,  is  the  number  36.  Thus,  100  lag  screws  of  this  size 
will  average  to  weigh  36  pounds,  and  one  such  screw  will  weigh 
0.36  pound ;  8  such  screws  will  weigh  0.36  X  8  =  2.88  pounds. 


SCREWS. 


409 


French  System  of  Standard  Threads. 

In  the  French  system  of  standard  screws  the  thread  has  an 
angle  of  60°,  with  flat  top  and  bottom.  (The  French  system  is 
in  this  respect  identically  the  same  as  the  United  States  stand- 
ard thread.)  The  pitch  and  the  diameter  are  given  in  millimeters 
(see  Table  No.  59).  The  form  of  thread  is  an  equilateral  tri- 
angle. The  diameter  of  tap  drill  (or  diameter  of  bolt  or  screw 
at  bottom  of  the  thread)  is  obtained  in  the  following  way  :  The 
height  of  an  equilateral  triangle  is  obtained  by  multiplying  its 
base  by  0.86603  (this  number  is  sin.  of  60°).  Thus,  assuming 
that  the  base  =  1,  and  taking  off  one-eighth  of  the  depth  at  top 
and  bottom,  that  is  reducing  the  depth  of  the  thread  one-fourth, 

0  86603 

the  remaining  depth  will  be  0.86603—  -^ =0.64952  ;  mul- 
tiplying this  by  2,  to  allow  for  the  depth  of  the  thread  on  both 
sides  of  the  screw,  the  constant  will  be  2  X  0.64952  =  1.29904, 
which  for  all  practical  purposes  may  be  reduced  to  1.3.  Thus, 
when  the  pitch  of  the  screw  is  1  millimeter,  the  diameter  of  the 
screw  at  the  bottom  of  the  thread  is  1.3  millimeters  less  than 
the  outside  diameter  of  the  screw.  Therefore,  the  diameter  of 
the  screw  at  the  bottom  of  the  thread  may  always  be  calculated 
by  the  simple  rule : 

Multiply  the  pitch  in  millimeters  by  1.3,  and  subtract  the 
product  from  the  outside  diameter  of  the  screw;  the  remainder 
is  the  diameter  of  the  screw  at  the  bottom  of  the  thread,  which 
is  the  same  as  the  diameter  of  the  tap  drill,  given  in  Table  No.  59. 

TABLE  No.  59.— French  Standard  Screws. 

(All  Dimensions  in  Millimeters). 


Diameter  of 
Screw. 

Pitch  of 
Screw. 

Diameter  of 
Tap  Drill. 

Diameter  of 
Screw. 

Pitch  of 
Screw. 

Diameter  of 
Tap  Drill. 

6 

1 

4.70 

36 

4 

30.80 

8 

1 

6.70 

42 

4.5 

36.15 

10 

1.5 

8.05 

48 

5 

41.50 

12 

1.5 

10.05 

56 

5.5 

48.85 

14 

2 

11.40 

64 

6 

56.20 

1(5 

2 

13.40 

72 

6.5 

63.55 

18 

2.5 

14.75 

80 

7 

70.90 

20 

25 

16.75 

88 

7.5 

78.25 

22 

2.5 

18.75 

96 

8 

85.60 

24 

8 

20.10 

106 

8.5 

94.95 

26 

3 

22.10 

116 

9 

104.30 

28 

3 

24.10 

12(5 

9.5 

1  13.65 

30 

3.5 

25.45 

ISO 

10 

123.00 

32 

3.5 

27.45 

148 

10.5 

134.35 

4io 


SCREWS. 


FIG. 


German  System  of  Standard  Threads. 

In  the  German  system  of  standard  threads  the  angle  is  53° 
7'  47/;.  The  reason  for  adopting  such  an  odd  angle  is  that  the 
form  of  thread  is  a  triangle,  having  its  base  equal  to  its  height, 

and  the  top  angle  of  such  a  triangle 
is  53°  7'  47".  The  thread  in  this 
system  is  also  made  flat  at  top  and 
bottom  equal  to  one-eighth  of  the 
pitch  (see  Fig.  4).  The  diameter 
of  the  screw  at  the  bottom  of  the 
thread  is,  in  this  system,  calculated 
by  this  rule : 

Multiply  the  pitch  in  millimeters  by  1.5,  subtract  this  pro- 
duct from  the  outside  diameter  of  the  screw  and  the  remainder 
is  the  diameter  of  the  screw  at  the  bottom  of  the  thread,  which 
is  the  same  as  the  diameter  of  the  tap  drill  given  in  Table  No.  60. 


TABLE  No.  60.— German  Standard  Screws. 

(All  Dimensions  in  Millimeters). 


•g 

•gj 

~ 

"0= 

~ 

H4 

S  £ 

'o  - 

Sft 

Sis 

"8  £ 

«Q 

>H          . 
<U     j£ 

"o  *' 

So 

.«C/3 

ll 

•F 

u  u 

11 

j£ 

ii 

«C/2 

JJ 

IE? 

ft 

1 

ft 

i 

0 

ft 

S^ 

* 

1 

0.25 

0.625 

5 

0.8 

3.8 

20 

2.4 

16.4 

1.2 

0.25 

0.825 

5.5 

0.9 

4.15 

22 

2.8 

17.8 

1.4 

0.3 

0.95 

6 

1 

4.5 

24 

2.8 

19.8 

1.7 

0.35 

1.175 

7 

1.1 

5.35 

26 

3.2 

21.2 

2 

0.4 

1.4 

8 

1.2 

6.2 

28 

3.2 

23.2 

2.3 

0.4 

1.7 

9 

1.3 

7.05 

30 

3.6 

24.6 

2.6 

0.45 

1.125 

10 

1.4 

7.9 

32 

3.6 

26.6 

3 

0.5 

2.25 

12 

1.6 

9.6 

36 

4 

30 

3.5 

0.6 

2.6 

14 

1.8 

11.3 

40 

4.4 

33.4 

4 

0.7 

2.95 

16 

2 

13 

4.5 

0.75 

3.375 

18 

2.2 

14.7 

International  Standard  for  Metric  Screw  Threads. 

An  international  standard  for  metric  screw  threads  was  dis- 
cussed at  a  congress  which  met  for  that  purpose  at  Zurich,  in 
October,  1898.  The  form  of  thread  adopted  is  based  on  the 
Sellers  thread,  which  it  will  be  remembered  has  the  shape  of  an 
equilateral  triangle  truncated  one-eighth  of  its  height  at  top  and 
bottom, 


SCREWS. 


411 


To  insure  Intel-changeability,  and  to  reduce  the  wear  on  taps 
and  dies,  the  congress  recommended  that  the  bottom  of  the 
thread  should  be  rounded  off  by  any  suitable  curve,  which  should 
not  deepen  the  cut  more  than  an  amount  equal  to  Vie  of  the 
pitch  beyond  the  standard  Sellers  type.  The  top  of  the  thread 
is  to  be  left  flat,  as  in  the  Sellers  system.  The  following  stand- 
ard sizes  and  pitches  were  decided  upon : 

TABLE  No.  6 1. —International  Standard  Thread. 


Diameter  in  Milli- 
meters. 

Pitch  in  Milli- 
meters. 

Diameter  in  Milli- 
meters. 

Pitch  in  Milli- 
meters. 

6  and  7 

1 

30  and  33 

3.5 

8  and  0 

1.25 

36           39 

4 

10  and  11 

1.5 

42           45 

4.5 

12 

1.75 

48           52 

5 

14  and  16 

2 

56           60 

5.5 

18,  20  and  22 

2.5 

64           68 

6 

24  and  27 

3 

72           76 

6.5 

To  Gear  a  Lathe  to  Cut  Metric  Thread  when  the 
Lead-Screw  is  in  Inches. 

Use  two  intermediate  gears,  one  having  100  teeth  and  the 
other  127  teeth,  fasten  these  two  gears  together  on  the  same  hub, 
and  gear  the  100-tooth  gear  into  the  gear  on  the  lead-screw 
and  the  127-tooth  gear  into  the  gear  on  the  stud  (see  Fig.  5). 

The  lathe  will  then  cut,  practi- 
cally, one-half  the  number  of  threads 
per  centimeter  that  it  originally  cut 
per  inch  with  a  common  intermediate 
gear.  For  instance,  the  stud  gear  has 
24  teeth,  the  screw  gear  has  48  teeth 
and  the  lead-screw  has  four  threads 
per  inch ;  the  lathe  will  then,  with  a 
common  intermediate  gear,  cut  eight 

F,G.  s.  •%^s^       threads  per  inch,  but  by  using  such  a 

double  intermediate  gear  as  is  shown 

in  Fig.  5  the  lathe  will  cut  four  threads  per  centimeter,  which  is 
the  same  as  one-fourth  times  10  and  equals  2  y2  millimeters  pitch, 
which  corresponds  to  a  metric  standard  screw  of  18  millimeters 
diameter. 

To  Calculate  the  Change  Gear  when  Cutting  fletric 
Screws  by  an  English  Lead- Screw. 

Divide  20  by  the  pitch  in  millimeters,  and  the  quotient  is  the 
corresponding  number  of  threads  per  inch  to  which  the  lathe 
must  be  geared. 


412 


SCREWS. 


EXAMPLE  1. 

To  gear  a  lathe  in  order  to  cut  a  metric  standard  screw 
24  millimeters  in  diameter  and  of  three  millimeters  pitch,  the 
lead-screw  on  the  lathe  having  four  threads  per  inch. 

Solution : 

Twenty  divided  by  three  gives  6%,  therefore  gear  the  lathe 
as  if  it  was  to  cut  6%  threads  per  inch  with  a  common  inter- 
mediate gear,  and  throw  in  the  special  intermediate  gear  as 
shown  in  the  cut,  and  the  lathe  will  cut  a  screw  of  three  milli- 
meters pitch.  The  gearing  is  easily  obtained,  thus: 

The  ratio  between  the  screw  gear  and  the  stud  gear  is  as  6%  to 
4,  which  is  the  same  as  20  to  12,  or,  reduced  to  its  lowest  terms,  5 
to  3.  Hence,  the  gears  may  have  any  number  of  teeth  providing 
the  ratio  is  5  to  8 ;  for  instance,  multiplying  by  9,  45  and  27 
could  be  used,  or,  multiplying  by  10,  50  and  30  could  be  used, 
etc. 

TABLE  No.  62.— How  to  Gear  a  Lathe  when  Cutting  Metric  Thread, 
Using  Inch-Divided  Lead-Screw  and  Intermediate  Gears,  as 
Shown  in  Fig.  5. 


h  i 

LEAD-SCREW   ON    LATHE. 

2 

3 

4 

5 

6 

10 

"If 
SHE 

Threads  per 
Inch. 

Threads  per 
Inch. 

Threads  per 
Inch. 

Thr 
per 

eads 
Inch. 

Threads 
per  Inch. 

Threads 
per  Inch. 

. 

. 

. 

, 

W  H  S 

T3  2 

w  S 

T3    2 

Es 

•58 

W    V 

T3  2 

w  S 

Js 

4>   2 

•a  2  £  ?, 

K  p^ 

3  Ml 

*-•    O/) 

3   bJD 

0   M 

3    bfl 

ui  tJO 

3  b£ 

i-t  bjo 

So 

IH    ftfj 

3  be 

5   M 

"T 

N 

C/3 

$ 

& 

£ 

* 

un 

£ 

Cfl 

en 

j 

24 

120 

20 

80 

24 

80 

20  !  40 

1.5 

24 

80 

30 

80 

36 

80 

30 

40 

2 

20 

100 

24 

80 

24 

60 

30 

60 

24 

40 

40 

40 

2.5 

20 

80 

24 

64 

24 

48 

30 

48 

30 

40 

50 

40 

3 

24 

80 

27 

60 

24 

40 

30 

40 

36 

40 

60 

40 

3.5 

28 

80 

21 

40 

28 

40 

35 

40 

42 

40 

70 

40 

4 

32 

80 

24 

40    i 

32 

40 

40 

40 

48 

40 

80 

40 

4.5 

27 

60 

27 

40 

36 

40 

45 

40 

54 

40 

5 

30 

60       30 

40 

40 

40 

50 

40 

60 

40 

5.5 

33 

60  i    33 

40 

44 

40 

55 

40 

66 

40 

6 

36 

60 

36 

40 

48 

40 

60 

40 

72 

40 

6.5 

39 

60 

39 

40 

52 

40 

65 

40 

7 

28 

40 

42 

40 

56 

40 

70 

40 

7.5 

30 

40 

45 

40 

60 

40 

8 

32 

40 

48 

40 

64 

40 

8.5 

34 

40 

51 

40 

9 

36 

40 

54 

40 

9.5 

38 

40 

57 

40 

10 

40 

40 

60 

40 

10.5 

42 

40 

11 

44 

40 

11.5 

46 

40 

12 

48 

40 

li 

NOTES    ON    HYDRAULICS.  413 

NOTES  ON  HYDRAULICS. 

Hydraulics  is  the  branch  of  engineering  treating  on  fluid  in 
motion,  especially  of  water,  its  action  in  rivers,  canals  and  pipes, 
the  work  of  machinery  for  raising  water,  the  work  of  water  as 
a  prime  mover,  etc. 

Pressure  of  Fluid  in  a  Vessel. 

When  fluid  is  kept  in  a  vessel  the  pressure  will  vary  directly 
as  the  perpendicular  height,  independent  of  the  shape  of  the 
vessel.  For  water,  the  pressure  is  0.434  pounds  per  square  inch, 
when  measured  one  foot  under  the  surface.  The  pressure  in 
pounds  per  square  inch  may,  therefore,  always  be  obtained  by 
multiplying  the  head  by  0.434.  The  head  corresponding  to  a 
given  pressure  is  obtained  by  either  dividing  by  0.434  or  multi- 
plying by  2.304. 

EXAMPLE. 

What  head  corresponds  to  a  pressure  of  80  pounds  per 
square  inch  ? 
Solution  : 

80  X  2.304  =  184  feet. 

Velocity  of  Efflux. 

The  velocity  of  the  efflux  from  a  hole  in  a  vessel  will  vary 
directly  as  the  square  root  of  the  vertical  distance  between  the 
hole  and  the  surface  of  the  water.  For  instance,  if  an  opening 
is  made  in  a  vessel  four  feet,  and  another  25  feet,  below  the  sur- 
face of  the  water,  and  the  vessel  is  kept  full,  the  theoretical 
velocity  of  the  efflux  will  be  nearly  16  feet  and  40  feet  per  second 
respectively,  friction  not  considered  ;  or,  in  other  words,  the 
velocity  will  be  as  2  to  5,  because  \^4  =  2  and  \^26  =  5. 

The  velocity  of  efflux  in  feet  per  second  may  always  be 
calculated  theoretically  by  the  formula  : 


v  =  8.02  X 

Constant  8.02  is  \S2g=  \/64.4,  and  v  —  velocity  of  efflux. 
h  =  Head  in  feet. 

Table  No.  63  gives  the  theoretical  velocity  of  efflux  and  the 
static  pressure  corresponding  to  different  heads,  and  is  calcu- 
lated by  the  following  formulas  : 


^  ~  644  *  =*  VTX  64.4  v  =  V  P  X  2.3  X  64.4 

v  =  V  P  X  148  P  =  h  X  0.434  h  —  P  X  2.3 


414 


NOTES    ON   HYDRAULICS. 


TABLE  No.  63.— Head,  Pressure,  and  Velocity  of  Efflux 
of  Water. 


Head  in 
Feet. 

Pressure  in 
Pounds  per 
Square  Inch. 

Velocity  in 
Feet  per 
Second. 

Head  in 
Feet. 

Pressure  in 
Pounds  per 
Square  Inch. 

Velocity  in 
Feet  per 
Second. 

h 

P 

i> 

h 

P 

V 

0.1 

0.0434 

2.54 

19 

8.246 

35 

0.2 

0.0868 

3.59 

20 

8.68 

35.9 

0.25 

0.1082 

4.01 

25 

10.82 

40.1 

0.3 

0.1302 

4.39 

30 

13.02 

43 

0.4 

0.1736 

5.07 

35 

15.19 

47.4 

0.5 

0.217 

5.67 

40 

17.36 

50.7 

0.6 

0.2604 

6.22 

45 

19.53 

53.8 

0.7 

0.3038 

6.71 

50 

21.7 

56.7 

0.75 

0.3255 

6.95 

55 

23.87 

59.5 

0.8 

0.3472 

7.18 

60 

26.04 

62.1 

0.9 

0.3906 

7.61 

65 

28.21 

64.7 

1 

0.434 

8.02 

70 

30.38 

67.1 

1.25 

0.5425 

8.95 

75 

32.55 

69.5 

1.5 

0.651 

9.83 

80 

34.72 

71.8 

1.75 

0.7595 

10.6 

85 

36.89 

73.9 

2 

0.868 

11.4 

90 

39.06 

76.1 

2.25 

0.9735 

12 

95 

41.23 

78.2 

2.5 

1.082 

12.6 

100 

43.4 

80.2 

2.75 

1.1905 

13.3 

110 

47.74 

84.2 

3 

1.302 

13.9 

120 

52.08 

87.68 

3.25 

1.4102 

14.4 

130 

56.78 

91.5 

3.5 

1.519 

15 

140 

61.06 

94.7 

3.75 

1.6375 

15.5 

150 

65.1 

98.3 

4 

1.736 

16 

160 

69.44 

101.2 

4.25 

1.8445 

16.5 

170 

73.78 

104.5 

4.5 

1.953 

17 

180 

78.12 

107.2 

4.75 

2.0615 

17.5 

190 

82.46 

110.4 

5 

2.17 

17.9 

200 

86.8 

113.5 

6 

2.604 

19.6 

225 

97.35 

120 

7 

3.038 

21.2 

250 

108.2 

126 

8 

3.472 

22.8 

275 

119.05 

133 

9 

3.906 

24.1 

300 

130.2 

139 

10 

4.34 

25.4 

325 

141.05 

144 

11 

4.774 

26.6 

350 

151.9 

150 

12 

5.208 

27.8 

375 

163.75 

155 

13 

5.678 

28.9 

400 

173.6 

160 

14 

6.106 

30 

425 

184.45 

165 

15 

6.51 

31.1 

450 

195.3 

170 

16 

6.944 

32.1 

475 

206.15 

174 

17 

7.378 

33.1 

500 

217 

179 

18 

7.812 

34 

550 

238.7 

188 

NOTES    ON    HYDRAULICS.  415 

Velocity  of  Water  in  Pipes. 

The  theoretical  velocity  of  water  discharged  from  a  pipe  is 
calculated  by  the  same  formula  as  is  used  in  calculating  veloci- 
ties of  falling  bodies.  (  See  page  277). 

v  =  \/2  g  h 

i>  =  Theoretical  velocity  of  efflux  per  second. 
h  =  Head. 

2g=  64.4  if  v  and  h  are  reckoned  in  feet. 
2  g  =  19.64  if  v  and  h  are  reckoned  in  meters. 
If  the  water,  besides  the  pressure  due  to  the  head,  is  also 
acted  upon  by  some  additional  pressure,  for  instance,  steam, 
the  theoretical  velocity  of  the  discharge  is  obtained  by   the 
formula, 


P  =  Pressure  in  pounds  per  square  inch. 

The  constant  0.434  is  used  because  a  column  of  water  one 
foot  high  will  exert  a  pressure  of  0.434  pounds  per  square  inch; 
thus,  by  dividing  by  0.434,  we  actually  convert  the  pressure  into 
its  corresponding  head  in  feet. 

All  other  quantities  in  this  formula  are,  of  course,  taken  in 
English  units. 

NOTE.  —  By  head  is  always  meant  the  vertical  height 
in  feet,  or  its  equivalent  in  pressure  expressed  in  feet.  Table 
No.  63  gives  the  theoretical  velocity  of  the  discharge  and  the 
pressure  corresponding  to  different  heads. 

The  theoretical  velocity  is  never  obtained  in  practice,  be- 
cause part  of  the  total  head  is  used  to  overcome  the  resistance 
at  the  entrance  of  the  pipe,  and  part  is  used  to  overcome 
the  frictional  resistance  to  the  flow  of  the  water  in  the  pipe. 
Thus,  only  a  part  of  the  total  head  is  left  to  give 
velocity  to  the  water,  therefore  the  velocity  of  the  water  at  dis- 
charge will  only  be  what  is  due  to  the  velocity  head,  after  deduc- 
tions are  made  for  resistance  at  the  entrance  and  for  friction  in 
the  pipes.  In  short  pipes,  the  resistance  at  the  entrance  to  the 
pipe  is  comparatively  the  larger  loss,  but  in  long  pipes  the  fric- 
tional resistance  is  the  larger. 

When  both  the  resistance  at  the  entrance  and  the  friction 
in  the  pipe  are  considered  the  formula  will  be  : 


•v  =  Velocity  of  discharge  in  feet  per  second. 
2  £-=64.4 

L  =  Length  of  pipe  in  feet. 

d=  Diameter  of  pipe  in  feet. 

f=  Coefficient  of  friction,  which  is  obtained  from  experi- 
ments, and  will  vary  according  to  conditions,  from  0.01  to  0.05. 
It  is  usually  in  approximate  calculations  taken  as  0.025. 


41 6  NOTES  ON  HYDRAULICS. 

EXAMPLE. 

Find  the  velocity  of  discharge  from  a  pipe  six  inches  in  diam- 
eter. The  head  is  16  feet  and  the  length  of  the  pipe  is  100 
feet,  and  coefficient  of  friction  0.025. 

Solution : 

(NOTE.    6  inches  =  0.5  foot.) 


64.4  X  16 


:.5  -f  0.025  X   -7TT 


6.5 
v  =  12.6  feet  per  second. 

In  Table  No.  64  the  quantity  of  water  discharged  per  min- 
ute by  a  pipe  six  inches  in  diameter,  when  the  velocity  is  one  foot 
per  second,  is  88.14  gallons.  Thus,  the  quantity  of  water  deliv- 
ered when  the  velocity  is  12.6  feet  per  second,  is  12.6  X  88.14  = 
1110.6  gallons  per  minute. 

When  the  length  of  the  pipe  is  more  than  4,000  diameters 
tne  velocity  of  the  water  may  be  calculated  by  the  formula, 


and  the  quantity  is  obtained  by  multiplying  the  velocity  by  the 
constants  given  in  Table  No.  64. 

EXAMPLE. 

Find  the  velocity  of  efflux  from  a  water  pipe  of  three 
inches  diameter  and  1200  feet  long,  having  a  head  of  six  feet, 
assuming  coefficient  of  friction  as  0.025. 

Solution  : 


0.25 


Discharge  in  gallons  per  minute : 

q  =  22.03  X  1.79  =  39.4  gallons  per  minute. 


NOTES   ON    HYDRAULICS. 


417 


TABLE  No.  64.— Quantity  of  Water  Discharged  Through 

Pipes  in  One  flinute,  when  Velocity  of  Efflux 

is  One  Foot  per  Second. 


-o 

* 

& 

•J  «  a 

r^- 

•s 

*0 

| 

o  *;  v 
'£*£, 

£60 

1-     • 

i 

v  ^ 

S  j 

^1     8 

^  rt  § 

P 

C  t"t 

^ 

c3|2| 

•i^l 

fl 

J! 

«£ 

s 

•|l°| 

IS 

c.s 

1  8 

5a 

Is 

•-S  ><% 

2^1 

1   0. 

1  u 

"la? 

J  w-2  o 

•ih-aS 

E.S. 

1  si, 

let 

J  «5>  o 

Jl"o 

Ss 

sfi 

£  a 

«fc  j>  t2 

«  o,i>  [2 

.£  £ 

'•J  c 

«5fc    rt(2 

cnj§  ^ 

c 

a 

c"* 

5 

S 

& 

a 

JS"" 

Q 

5 

% 

0.0104  0.00008 

0.0048 

0.036 

20 

1.6666 

2.182 

130.90 

979 

}|  ;  0.0208  0.00033  j     0.0198!     0.150 

21 

1.7500 

2.405 

144.32 

1079 

%    0.03120.00076!     0.0456      0.342 

22 

1.8333    2.640 

158.39 

1185 

X    0.04160.00136!     0.0816J     0.612 

23 

1.9166    2.885 

173.11 

1294 

% 

0.0624  0.00306J     0.1836      1.380: 

24 

2.000 

3.142 

188.50 

1410 

1 

0.0833  0.00545 

0.3272 

2.448 

25 

2.0833!  3.409 

204.53 

1530 

lj 

0.1042  0.00852 

0.5094 

3.828 

26 

2.1667 

3.687 

221.22 

1655 

0.12500.01227 

0.7362 

5.508 

27 

2.2500 

3.976 

238.56 

1784 

2s 

0.1667  0.0218 

1.309 

9.792 

28 

2.3333 

4.276 

256.56 

1919 

2i 

0.2083  0.0341 

2.045 

15.30 

29 

2.4166 

4.578 

275.22 

2058 

3 

0.25000.0491 

2.945 

22.03 

30 

2.5000 

4.909 

294.52 

2203 

^1 

0.29110.0668 

4.008 

29.99 

31 

2.5822 

5.241 

314.49 

2352 

4 

0.3333  0.0873 

5.238 

39.17 

32 

2.6667 

5.585 

335.10 

2506 

41 

0.37500.1104 

6.626 

49.58 

33 

2.7500 

5.939 

356.37 

2666 

5 

0.4166  0.1364 

8.181 

61.20 

34 

2.8333 

6.305 

378.302829 

6 

0.5000  0.1963 

11.781 

88.14 

35 

2.9166 

6.681 

400.88  '2999 

7 

0.5822  0.2673 

16.035 

119.9 

36 

3.0000 

7.069 

424.14  3173 

8 

0.6667  0.3491 

20.914 

156.7 

37 

3.0833 

7.467 

448.02  3351 

9  (0.75000.4418 

26.507 

198.3 

38 

3.1667 

7.876 

472.56 

3535 

10    0.83330.5454 

32.725 

244.8 

39 

3.2500 

8.296 

497.75  3724 

11 

0.9166!0.6597 

39.597 

296.2 

40 

3.3333 

8.727 

523.60 

3918 

12     1.00000.7854 

47.124 

352.5 

41 

3.4166 

9.168 

550.11 

4115 

13    1.0833 

0.9218 

55.305 

413.7 

42 

3.5000 

9.621 

577.27 

4318 

14     1.1667 

1.069 

64.141 

479.8 

43 

3.5822 

10.085 

605.09 

4526 

15 

1.2500 

1.227 

73.631 

550.8 

44 

3.6667 

10.559 

633.56 

4739 

16 

1.3333 

1.396 

83.776 

626.4 

45 

3.7500!  11.  045 

662.68 

4961 

17 

1.4166 

1.576 

94.575 

707.4 

46 

3.8333  11.541 

692.46 

5180 

18 

1.5000 

1.768 

106.03 

793.2 

47 

3.9166 

12.048 

722.90 

5408 

19 

1.5822 

1.969 

118.14 

883.8 

48 

4 

12.566 

753.98 

5640 

41 8  NOTES   ON   STEAM. 


NOTES  ON  STEAM. 

When  water  is  heated  and  converted  into  steam  of  atmos- 
pheric pressure,  one  cubic  foot  of  water  will  make  1646  cubic 
feet  of  steam.  (The  common  expression  that  **  a  cubic  inch  of 
water  makes  a  cubic  foot  of  steam  "  is  not  strictly  correct,  as  a 
cubic  foot  contains  1728  cubic  inches.) 

The  specific  gravity  of  steam  at  atmospheric  pressure,  when 

compared  with  water  is,  therefore.  -         =  0.000608. 


The  weight  of  one  cubic  foot  of  steam  at  atmospheric 
pressure  will,  therefore,  be  0.000608  X  62.5  =  0.038  pounds.  At 
any  other  pressure  the  weight  per  cubic  foot  of  steam  is  given 
in  Table  No.  65. 

Saturated  steam  is  steam  at  the  temperature  of  the  boiling 
point  which  corresponds  to  its  pressure.  Saturated  steam  does 
not  need  to  be  wet  steam,  as  the  word  saturated  does  not  mean 
that  the  steam  is  saturated  with  water,  but  it  means  that  it  is 
saturated  with  heat  ;  that  is  to  say  :  the  temperature  under  the 
given  pressure  cannot  possibly  be  any  higher  as  long  as  the 
steam  is  in  contact  with  water,  because  if  more  heat  is  added 
more  water  will  be  evaporated,  and  if  the  volume  is  kept  con- 
stant, as  in  a  steam  boiler,  both  the  pressure  and  temperature 
will  increase  simultaneously. 

High  pressure  steam  is  steam  the  pressure  of  which  greatly 
exceeds  the  pressure  of  the  atmosphere. 

Low  pressure  steam  is  steam  the  pressure  of  which  is  less 
than  the  atmosphere,  and  also  steam  having  a  pressure  equal 
to,  or  not  greatly  above,  the  atmospheric  pressure. 

Wet  steam  is  steam  which  contains  water  held  in  suspen- 
sion mechanically. 

Dry  steam  is  steam  which  does  not  contain  water  held  in 
suspension  mechanically. 

Super-heated  steam  is  steam  which  is  heated  to  a  tempera- 
ture higher  than  the  boiling  point  corresponding  to  its  pressure. 
It  cannot  exist  in  contact  with  water,  nor  contain  water,  and 
resembles  a  perfect  gas.  Vertical  boilers  with  tubes  through 
the  steam  space  (such  as  the  Manning  boiler)  give  slightly  super- 
heated steam  ;  but  if  steam  is  to  be  super-heated  to  any  consid- 
erable extent  it  must  be  passed  through  a  super-heater,  which 
usually  is  in  the  form  of  a  coil  of  pipes  subjected  to  the  hot 
gases  in  the  uptake  from  the  boiler. 

The  sensible  heat  of  steam  is  the  temperature  which  can 
be  measured  by  a  thermometer. 

The  latent  heat  of  steam  is  that  heat  which  is  absorbed 
when  water  of  any  given  temperature  is  changed  into  steam  of 
the  same  temperature. 

When  water  is  evaporated  under  pressure  the  sensible  heat 
will  increase  and  the  latent  heat  will  decrease.  For  instance, 
at  atmospheric  pressure  the  sensible  heat  is  212  degrees,  and  the 
latent  heat  of  evaporation  is  966  B.  T.  U.,  but  at  100  pounds 


NOTES  ON   STEAM. 


419 


absolute  pressure  the  sensible  heat  is  327.9  degrees,  while  the 
latent  heat  of  evaporation  is  only  883.1  B.  T.  U.  (See  steam 
table,  No.  65.) 

TABLE  No.  65.— Properties  of  Saturated  Steam. 


ill. 

fM 

W 

ffrf 

!«!& 
l^fi 

Total  Heat  in 
B.  T.  U.  per 
Pound  of  Steam 
from  Water  at 
32  Degrees  F. 

&3 

3HJ 

g*w^ 

3W.S£ 

.a!~o  . 

i!*S 

III* 

Slli 

.Sfa  o% 
£S3W 
*&* 

Cubic  Feet  of 
Steam  from  1 
Cubic  Foot  of 
Water  at  62 
Degrees  F. 

i 

102.1 

1112.5 

1042.9 

330.36 

0.0030 

20800 

2 

126.3 

1119.7 

1025.8 

172.80 

0.0058 

10760 

3 

141.6 

1124.6 

1014 

117.52 

0.0085 

7344 

4 

153.1 

1128.1 

1006.8 

89.36 

0.0112 

5573 

5 

162.3 

1130.9 

1000.3 

72.80 

0.0138 

4524 

6 

170.2 

1133.3 

995 

61.52 

0.0163 

3813 

7 

176.9 

1135.3 

990 

52.62 

0.0189 

3298 

8 

182.9 

1137.2 

985.7 

46.66 

0.0214 

2909 

9 

188.3 

1138.8 

982.4 

41.79 

0.0239 

2604 

10 

193.3 

1140.3 

978.4 

37.84 

0.0264 

2358 

11 

197.8 

1141.7 

975.3 

34.63 

0.0289 

2157 

12 

202 

1143 

972.2 

31.88 

0.0314 

1980 

13 

205.9 

1144.2 

970 

29.57 

0.0338 

1842 

14 

209.6 

1145.3 

968 

27.61 

0.0362 

1720 

14.7 

212 

1146.1 

966 

26.36 

0.0380 

1646 

15 

213.1 

1146.4 

964.3 

25.85 

0.0387 

1610 

16 

216.3 

1147.7 

962.6 

24.32 

0.0411 

1515 

17 

219.6 

1148.3 

960.4 

22.96 

0.0435 

1431 

18 

222.4 

1149.2 

957.7 

21.78 

0.0459 

1357 

19 

225.:} 

1150.1 

956.3 

20.70 

0.0483 

1290 

20 

228 

1150.9 

952.8 

19.72 

0.0507 

1229 

25 

240 

1154.6 

945.3 

15.99 

0.0625 

996 

30 

250 

1157.8 

937.9 

13.46 

0.0743 

838 

86 

259.3 

1160.5 

931.6 

11.65 

0.0858 

726 

40 

267.3 

1162.9 

926 

10.27 

0.0974 

640 

45 

274.4 

1165.1 

920.9 

9.18 

0.1089 

572 

50 

281 

1167.1 

916.3 

8.11 

0.1202 

518 

55 

287.1 

1169 

912 

7.61 

0.1314 

474 

60 

292.7 

1170.7 

908 

7.01 

0.1425 

437 

65 

298 

1172.3 

904.2 

6.49 

0  1538 

405 

70 

302.9 

1173.8 

900.8 

6.07 

0.1648 

378 

75 

307.5 

1175.2 

897.5 

5.68 

0.1759 

353 

80 

312 

1170.5 

894.3 

5.35 

0.1869 

333 

85 

316.1 

1177.9 

891.4 

5.05 

0.1980 

314 

90 

320.2 

1179.1 

888.5 

4.79 

0.2089 

298 

95 

324 

1180.3 

885.8 

4.55 

0.2198 

283 

100 

327.9 

1181.4 

883.1 

4.33 

0.2307 

270 

105 

331.3 

1182.4 

881.7 

4.14 

0.2414 

257 

420 


NOTES   ON   STEAM. 


TABLE   No.  65.  — (Continued). 


Absolute  Pres- 
sure in  Pounds 
per  Square 
Inch. 

1  =-s^* 

fP! 

Total  Heat  in 
B.  T.  U.  per 
Pound  of  Steam 
from  Water  at 
32  Degrees  F. 

Latent  Heat  of 
Evaporation 
in  B.  T.  U.  per 
Pound  of 
Steam. 

4*. 

S    WT3    6 

IS|l 

>-§* 
u 

Weight  in 
Pounds  per 
Cubic  Foot  of 
Steam. 

Cubic  Feet  of 
Steam  from  1 
Cubic  Foot  of 
Water  at  62 
Degrees  F. 

110 

334.6 

1183.5 

878.3 

3.97 

0.2521 

247 

115 

338 

1184.5 

875.9 

3.80 

0.2628 

237 

120 

341.1 

1185.4 

873.7 

3.65 

0.2738 

227 

125 

344.2 

1186.4 

871.5 

3.51 

0.2845 

219 

130 

347.2 

1187.3 

869.4 

3.38 

0.2955 

211 

135 

350.1 

1188.2 

867.4 

3.27 

0.3060 

203 

140 

352.9 

1189 

865.4 

3.16 

0.3162 

197 

145 

355.6 

1189.9 

863.5 

3.06 

0.3273 

190 

150 

358.3 

1190.7 

861.5 

2.96 

0.3377 

184 

160 

363.4 

1192.2 

857.9 

2.79 

0.3590 

174 

170 

368.2 

1193.7 

854.5 

2.63 

0.3798 

164 

180 

372.9 

1195.1 

851.3 

2.49 

0.4009 

155 

190 

377.5 

1196.5 

848 

2.37 

0.4222 

148 

200 

381.7 

1197.8 

845 

2.26 

0.4431 

141 

In  the  preceding  table  the  first  column  gives  the  absolute 
pressure,  which  is  gage  pressure  plus  14.7  pounds,  or,  for 
ordinary  practice,  reckon  as  15  pounds.  For  instance,  when 
the  gage  pressure  is  80  pounds  per  square  inch,  the  correspond- 
ing absolute  pressure  is,  for  all  practical  purposes,  95  pounds 
per  square  inch,  and  the  corresponding  temperature  is  given  in 
the  second  column  in  the  table  to  be  324  degrees  Fahr. 

The  total  number  of  British  thermal  units  (B.  T.  U.)  re- 
quired to  convert  each  pound  of  water  from  32  degrees  Fahr. 
into  steam  of  any  given  pressure  is  given  in  the  third  column. 
For  instance,  each  pound  of  water  of  32  degrees  converted 
into  steam  of  95  pounds  per  square  inch  absolute  pressure  has 
received  1180.3  B.  T.  U. 

The  fourth  column  gives  the  number  of  British  thermal 
units  (B.  T.  U.)  of  heat  required  to  change  one  pound  of  water 
of  the  temperature  given  in  the  second  column  into  steam  of 
the  same  temperature ;  which  also  is  the  number  of  heat 
units  given  up  by  one  pound  of  steam  when  it  is  condensed  to 
water  of  the  same  temperature  as  the  temperature  of  the  steam 
with  which  it  is  in  contact.  For  instance,  the  table  gives  the 
latent  heat  of  evaporation  of  steam  at  95  pounds  absolute  pres- 
sure to  be  885.8  B.  T.  U.;  therefore,  if  steam  of  95  pounds  pres- 
sure per  square  inch  is  condensing  into  water  in  a  steam  pipe 
where  steam  and  water  are  in  contact,  so  that  the  temperature 
cannot  drop  below  that  due  to  the  pressure,  and  the  pressure  is 


NOTES    ON    STEAM.  421 

maintained  at  95  pounds  per  square  inch,  the  temperature  of 
the  water  from  the  condensed  steam  will  be  324  degrees,  the 
oame  as  the  temperature  of  the  steam,  but  each  pound  of  steam 
as  it  is  condensing  will  give  out  885.8  British  thermal  units  of 
heat. 

The  fifth  column  gives  the  number  of  cubic  feet  of  satu- 
rated steam  which  will  weigh  one  pound  at  the  given  pressure 
and  temperature.  The  sixth  column  gives  the  weight  of  one 
cubic  foot  of  saturated  steam  of  corresponding  given  tempera- 
ture. For  instance,  one  cubic  foot  of  steam  at  95  pounds  per 
square  inch  absolute  pressure  will  weigh  0.2198  pounds,  and  100 
cubic  feet  of  steam  of  95  pounds  per  square  inch  absolute  pres- 
sure will  weigh  100  X  0.2198  =  21.98  pounds.  In  other  words,  it 
will  take  0.2198  pounds  of  water  to  give  one  cubic  foot  of  steam 
at  95  pounds  absolute  pressure,  and  it  will  require  21.98  pounds 
of  water  to  make  100  cubic  feet  of  steam  of  95  pounds  absolute 
pressure. 

The  seventh  or  last  column  gives  the  relative  volume  of 
steam  at  the  given  pressure  as  compared  with  water  at  32 
degrees  F.  For  instance,  one  cubic  foot  of  water  will  give 
1046  cubic  feet  of  steam  at  atmospheric  pressure,  but  one  cubic 
foot  of  water  gives  only  219  cubic  feet  of  steam  at  125  pounds 
absolute  pressure. 

Steam  Heating. 

In  the  ordinary  practice  of  heating  buildings  by  direct 
radiation  the  quantity  of  heat  given  off  by  the  radiators  or  steam 
pipes  will  vary  from  1  ^  to  3  heat  units  per  hour  per  square  foot 
of  radiating  surface  for  each  degree  of  difference  in  tempera- 
ture; an  average  of  from  2  to  2^  is  a  fair  estimate. 

One  pound  of  steam  at  about  atmospheric  pressure  contains 
1146  heat  units,  and  if  the  temperature  in  the  room  is  to  be 
maintained  at  70°,  while  the  temperature  of  the  pipes  is 
212°,  the  difference  in  temperature  will  be  142  degrees.  Multi- 
plying this  by  2#,  the  emission  of  heat  will  be  229^  heat  units 
per  hour  per  square  foot  of  radiating  surface.  Dividing  229)4 
by  1146  gives  0.2  pounds  of  steam  condensed  per  hour,  per  square 
foot  of  radiating  surface.  From  this  may  be  estimated  the  re- 
quired size  of  boiler,  as  the  boiler  must  always  be  capable  of 
generating  as  much  steam  as  the  radiators  are  condensing.  A 
rule  frequently  given  is  to  have  one  square  foot  of  heating  sur- 
face in  the  boiler  for  every  8  to  10  square  feet  of  radiating  sur- 
face and  one  square  foot  of  grate  surface  for  every  350  to  500 
square  feet  of  radiating  surface. 

One  pound  of  coal  is  required  per  hour  per  30  to  40  square 
feet  of  radiating  surface. 

When  steam  is  used  for  heating  dwelling-houses,  one  square 
foot  of  radiating  surface  is  required  per  40  to  80  cubic  feet  of 
space,  according  to  location,  number  of  windows,  etc.  As  a 


422  NOTES   ON   STEAM. 

general  rule,  one  square  foot  of  radiating  surface  is  sufficient 
for  heating  40  to  60  cubic  feet  of  air  in  outer  or  front  rooms, 
and  80  to  100  cubic  feet  in  inner  rooms.  The  following  rule  may 
be  used  as  a  guide  for  different  conditions :  One  square  foot  of 
radiating  surface  is  sufficient  for'  heating  60  to  80  cubic  feet  of 
space  in  dwellings,  schools  and  offices ;  75  to  100  cubic  feet  of 
space  in  halls,  store  houses  and  factories ;  150  to  200  cubic  feet 
of  space  in  churches  and  large  auditoriums. 

In  heating  mills  1^-inch  steam  pipes  are  generally  used, 
and  one  foot  of  pipe  is  allowed  per  90  cubic  feet  of  space  to  be 
heated. 

Value  of  Low  Pressure  Steam  for  Heating  Purposes. 

When  steam  at  atmospheric  pressure  is  condensed  into 
water  at  a  temperature  of  212°,  each  pound  of  steam  gives  up 
966  B.  T.  U.  of  heat,  but  if  steam  of  100  pounds  gage  pressure 
(115  pounds  absolute)  is  condensed  into  water  at  212  degrees, 
each  pound  of  steam  must  give  up  1004  B.  T.  U.,  which  is  only 
38  heat  units  more  than  steam  of  atmospheric  pressure.  Hence 
it  is  evident  that  for  heating  purposes  there  is  no  advantage  in 
using  steam  of  high  pressure ;  one  pound  of  exhaust  steam,  only 
a  pound  or  two  over  atmospheric  pressure,  is  almost  as  valuable 
an  agent  for  heating  purposes  as  live  steam  at  100  pounds  pres- 
sure direct  from  the  boiler. 

Hot  Water  Heating  in  Dwelling  Houses. 

One  square  foot  of  heating  surface  is  required  per  30  to  60 
cubic  feet  of  space  heated. 

Quantity  of   Water  Required  to  Make  any   Quantity  of 
Steam  at  any  Pressure. 

The  weight  of  water  required  to  make  one  cubic  foot  of 
steam  at  any  pressure  is  the  same  as  the  weight  of  one  cubic 
foot  of  steam  as  given  in  the  sixth  column  in  Table  No.  65. 

Therefore,  the  weight  of  water  is  obtained  by  multiplying 
the  number  of  cubic  feet  of  steam  required  by  the  weight  of  one 
cubic  foot,  as  given  in  the  table. 

EXAMPLE. 

How  much  water  will  it  take  to  make  300  cubic  feet  of  steam 
at  100  pounds  absolute  pressure  ? 

Solution : 

One  cubic  foot  of  steam  at  100  pounds  pressure  is  given  in 
the  table  as  weighing  0.2307  pounds,  therefore  300  cubic  feet 
will  weigh  300  X  0.2307  =  69.21  pounds  of  water. 

One  cubic  foot  of  water  may,  for  any  practical  purpose,  be 
reckoned  to  weigh  62  ^  pounds  and  one  gallon  of  water  may  be 


NOTES   ON   STEAM.  423 

taken  as  &fo  pounds,  Therefore  69.21  pounds  divided  by  62.5 
gives  1.1  cubic  feet,  or  69.21  pounds  divided  by  8.3  gives  8.34 
gallons. 

At  atmospheric  pressure  one  cubic  foot  of  steam  has  nearly 
the  weight  of  one  cubic  inch  of  water,  and  the  weight  increases 
very  nearly  as  the  pressure ;  therefore,  for  an  approximate  estima- 
tion, if  no  steam  tables  are  at  hand,  it  is  well  to  remember  the 
rule: 

Multiply  the  number  of  cubic  feet  of  steam  by  the  absolute 
pressure  in  atmospheres,  and  the  product  is  the  number  of  cubic 
inches  of  water  required  to  give  the  steam. 

NOTE. — In  all  such  calculations  for  practical  purposes,  a 
liberal  allowance  must  be  made  for  loss  and  leakage. 

Weight  of  Water  Required  to  Condense  One 
Pound  of  Steam. 

The  following  formula  gives  the  theoretical  amount  of  water 
required  to  condense  one  pound  of  steam : 


/2  t\ 

W  =  Weight  of  water  required  per  pound  of  steam  con- 
densed. 

//  =  Number  of  heat  units  above  32°  in  one  pound  of  steam 
at  the  pressure  of  exhaust.  This  temperature  is 
obtained  from  Table  No.  65. 

t\  =  Temperature  of  water  when  entering  the  condenser. 
/2  =  Temperature  of  water  when  leaving  the  condenser. 
/3  =  Temperature  of  the  condensed  steam  when  leaving 
the  condenser  and  entering  the  air-pump. 

EXAMPLE. 

Steam  of  four  pounds  absolute  pressure  is  exhausted  into  a 
surface  condenser.  The  temperature  of  the  condensed  steam 
when  leaving  the  condenser  and  entering  the  air-pump  is  120°. 

The  temperature  of  the  cold  water  when  entering  the  con- 
denser is  65°. 

The  temperature  when  leaving  the  condenser  is  105°. 
How  many  pounds  of  condensing  water  is  needed  per  pound  of 
steam  condensed  ? 

NOTE. — In  the  steam  table,  page  419,  the  total  number  of 
heat  units  above  32°  per  pound  of  steam  of  four  pounds  absolute 
pressure  is  given  as  1128. 

Solution : 

_  1128  +  32  —  120 
105  —  65 

1040 
IV  —  —   --  =  26  pounds  of  water  per  pound  of  steam. 

40 


424  NOTES   ON   STEAM. 

In  a  jet  condenser  the  steam  and  the  water  are  mixed  to- 
gether, and,  therefore,  the  condensed  steam  and  the  water  when 
leaving  the  condenser  are  of  equal  temperature,  and  the  formula 
will  change  to 


/2  =  Temperature  of  mixture. 

t\  =  Temperature  of  water  when  entering  condenser. 
The  other  letters  have  the  same  meaning  as  in  the  previous 
formula. 

EXAMPLE. 

Steam  of  three  pounds  absolute  pressure  in  exhausted  into 
a  jet  condenser.  The  temperature  of  the  cold  water  entering  is 
60°.  The  temperature  of  the  mixture  leaving  the  condenser  is 
110°.  How  many  pounds  of  water  are  needed  per  pound  of 
steam  condensed? 

NOTE. — In  the  steam  table,  page  419,  the  total  number  of 
heat  units  above  32°  per  pound  of  steam  of  three  pounds  pres- 
sure is  given  as  1124.0  or,  for  convenience,  say  1125. 

Solution : 

1125  +  32  —  110 

lA/    — *    

110  —  CO 
W  =  =  20.1  pounds. 

Weight  of  Steam  Required  to  Boil  Water. 

An  approximate  rule  is  to  allow  that  one  pound  of  steam  is 
condensed  for  every  five  pounds  of  water  to  be  heated  to  the 
boiling  point. 

It  does  not  make  much  difference  about  the  pressure  of  the 
steam,  as  long  as  it  is  a  few  pounds  above  atmospheric  pres- 
sure ;  for  instance,  one  pound  of  steam  at  10  pounds  gage  pres- 
sure when  condensed  into  water  at  212°  will  give  up  973  heat 
units,  and  steam  of  100  pounds  gage  pressure  will  give  up  1003 
heat  units — a  difference  of  only  30  heat  units  in  steam  of  10 
pounds  gage  pressure  and  steam  of  100  pounds  gage  pressure. 

More  correctly,  the  weight  of  steam  required  to  boil  one 
pound  of  water  at  212°  may  be  calculated  by  the  formula, 
_  212  —  A 
~  //—ISO 

And  the  weight  of  steam  required  to  heat  one  pound  of  water  to 
any  temperature  is  obtained  by  the  formula, 

*~  H  +  32  —  /2 


NOTES    ON   STEAM.  425 

x  =  Weight  of  steam  required. 
H  =  Number  of  heat  units  above  32°  in  one  pound  of  steam, 

as  given  in  Table  No.  65. 
/i  =  Temperature  of  water  before  heating. 
t%  =  Temperature  of  water  after  heating. 


Expansion  of  Steam  in  Steam  Engines. 

When  steam  is  expanded  without  doing  work  and  prac- 
tically without  losing  heat  by  radiation,  it  will  become  super- 
heated, but  if  it  is  doing  work,  as  in  a  steam  engine,  it  will  lose 
heat  during  expansion. 

According  to  the  best  authorities,  the  pressure  varies  in- 

jo  <  versely  as  the  ^power  of  the  volume,  if  heat  is  neither  added 

'nor  taken* awayT>y  any  outside  source  during  the  time  the  steam 

is  being  expanded  in  the  steam  engine  cylinder.     This  is  called 

adiabatic  expansion  of  steam. 

The  pressure  varies  inversely  as  the  1Vio  power  of  the  vol- 
ume, if  the  steam  is  kept  dry  at  the  temperature  of  saturation, 
during  expansion,  by  means  of  a  steam  jacket  outside  the 
cylinder. 

When  the  pressure  is  considered  to  vary  inversely  as  the 
volume  it  is  called  isothermal  expansion. 

The  isothermal  curve  is  not  exactly  the  correct  curve  to 
represent  the  expansion  of  steam,  but  it  is  the  theoretical  curve 
usually  drawn  on  the  indicator  diagram,  because  it  is  so  easy  to 
handle  and  is  also  very  nearly  correct. 

The  following  formula  gives  the  mean  effective  pressure 
according  to  isothermal  expansion. 


M.  E.  P.  = 


+  hyp.log.r 


p.log.r\  __ 


Absolute  terminal  pressure  =  PI  X  — • 

Pi  =  Absolute  initial  pressure. 
r  =  Ratio  of  expansion. 

PZ  =  Absolute  back  pressure. 
M.  E.  P.  —  Mean  effective  pressure. 
Hyp.  log.  (hyperbolic  logarithm),  see  page  126. 

Table  No.  66  gives  the  terminal  and  the  mean  effective 
pressure  of  steam  expanded  under  any  of  these  three  different 
conditions. 


426 


NOTES    ON    STEAM. 


TABLE   No.  66.— Constants   for   Calculating   Mean 
Terminal  Pressure  of  Expanding  Steam. 


and 


• 

Kept  Dry  at  Tem- 

V 

,3 

0 

At  Constant  Tem- 
perature. 
(Isothermal    Expan- 

perature   of    Satura- 
tion. 
(Expansion  in  a 

Condensing  by  Work- 
ing in  a  Cylinder. 
(Adiabatic  Expan- 

w 

C/2 

sion). 

Steam  Jacketed  Cy- 

sion). 

D 

V 

linder)  . 

'5 

03 

jv 

§ 

V 

g 

9 

I 

i 

3 

0 

2 

I 

1*^ 

ti  <N 

3     ^_~/ 

v. 

1  a 

=  H^  v 

£ 

*«S 

^  i-<|  k 

<8  ti;  v 

PH   o 

$       CO 

PH   £ 

i  s-x 

1 

ji! 

3« 

1 

ii 

IS' 

&           1 

|5- 

1  1 

CJ 

55 

H 

*.'-. 

H 

£ 

H 

£ 

% 

1% 

0.750 

0.966 

0.737 

0.964 

0.726 

0.962 

7/10 

1% 

0.700 

0.950 

0.685 

0.947 

0.673 

0.944 

X 

1% 

0.667 

0.937 

0.650 

0.933 

0.638 

0.930 

% 

1% 

0.625 

0.919 

0.607 

0.914 

0.593 

0.911 

6/10 

1% 

0.600 

0.906       0.581 

0.900 

0.567 

0.897 

X 

2 

0.500 

0.846       0.479 

0.839 

0.463 

0.833 

4/10 

2% 

0.400 

0.766    ii  0.378 

0.756 

0.361 

0.748 

% 

2% 

0.375 

0.743       0.353 

0.732 

0.336 

0.723 

% 

3 

0.333 

0.700 

0.311 

0.688 

0.295 

0.678 

3/10 

3% 

0.300 

0.662       0.278 

0.648 

0.262 

0.637 

* 

4 

0.250 

0.596 

0.229 

0.582 

0.214 

0.571 

5 

0.200 

0.522    i|  0.181 

0.506 

0.167 

0.495 

1/Q 

6 

0.1667 

0.465     ,  0.149 

0.449 

0.137 

0.437 

y? 

7 

0.1428 

0.421 

0.127 

0.405 

0.115 

0.393 

X 

8 

0.125 

0.385 

0.110 

0.369 

0.0992 

0.357 

M> 

9 

0.1111 

0.355 

0.0968 

0.340 

0.0862 

0.327 

Ho 

10 

0.1000 

0.330 

0.0865 

0.315 

0.0774 

0.303 

Hi 

11 

0.0909 

0.309 

0.0782 

0.293 

0.0696 

0.282 

Hi 

12 

0.0833 

0.290 

0.0713 

0.275 

0.0631 

0.264 

Vis 

13 

0.0769 

0.274 

0.0655 

0.259 

0.0578 

0.248 

M.4 

14 

0.0714 

0.260 

0.0605 

0.245 

0.0533 

0.234 

¥15 

15 

0.0667 

0.247 

0.0563 

0.232 

0.0494 

0.222 

16 

0.0625 

0.236 

0.0526 

0.221 

0.0459 

0.211 

^7 

17 

0.0588 

0.225 

0.0493 

0.211 

0.0429 

0.201 

%8 

18 

0.0556 

0.216 

0.0463 

0.202 

0.0403 

0.192 

H9 

19 

0.0526 

0.208 

0.0438 

0.193 

0.0379 

0.184 

20 

0.0500 

0.200 

0.0415 

0.185 

0.0359 

0.177 

NOTES   ON   STEAM.  4.27 

To  Find  the  Mean  Effective  Pressure  by  the 
Preceding  Table. 

Find  the  constant  in  the  column  corresponding  to  the  con- 
ditions of  expansion,  and  to  the  given  cut-off.  Multiply  this  by 
the  absolute  initial  pressure,  and  the  product  is  the  average 
pressure.  Subtract  the  back  pressure  and  the  remainder  is  the 
mean  effective  pressure. 

EXAMPLE. 

Find  the  mean  effective  pressure  for  isothermal  expansion 
when  the  engine  is  cutting  off  at  one-quarter  stroke.  The  initial 
pressure  is  90  pounds  absolute.  The  absolute  back  pressure  is 
18  pounds. 

Solution  : 

M.  E.  P.  =  90  X  0.596  —  18  =  53.64  —  18  —  35.64  pounds. 

NOTE. — All  such  calculations  must  be  made  from  absolute 
pressure  (not  gage  pressure),  and  when  determining  the  cut-off 
the  clearance  must  be  considered. 

Clearance. 

The  clearance  of  an  engine  is  usually  expressed  as  a  per- 
centage of  the  piston  displacement.  The  space  between  the 
piston  and  the  cylinder  head  at  the  end  of  the  stroke,  also  the 
cavities  due  to  the  steam  ports,  must  be  included  in  considering 
clearance. 

In  high-class  Corliss  engines  the  clearance  does  not  exceed 
2  l/z  to  5  per  cent.,  but  in  common  slide-valve  engines  the  clear- 
ance may  go  as  high  as  5  to  15  per  cent.  When  clearance  is 
taken  into  account  the  actual  ratio  of  expansion  is 


R  =  Actual  ratio  of  expansion. 
r  =  Nominal  ratio  of  expansion. 

c  =  Clearance,  expressed  as  a  fractional  part  of  the  length 
of  the  stroke. 

EXAMPLE. 

The  nominal  ratio  of  expansion  is  4,  and  the  clearance  is 
5  per  cent.     What  is  the  actual  ratio  of  expansion? 

NOTE. — 5  per  cent,  is  %oo  =  Vso  =  0.05  of  the  stroke. 
Solution : 
R  _    1  +  0.05 

-±-  +  0.05 
R  =  -j-£-  =  3.5  =  Actual  ratio  of  expansion. 

O.o 


428 


SHOP   NOTES. 


SHOP    NOTES. 

Weight  of  a  Grindstone. 

Multiply  the  constant  0.064  by  the  square  of  the  diameter 
in  inches  and  this  product  by  the  thickness  in  inches;  the  result 
is  the  weight  of  the  grindstone  in  pounds. 

EXAMPLE. 

Find  the  weight  of  a  grindstone  30  inches  in  diameter  and 
six  inches  thick. 

Solution : 

Weight  =  0.064  X  30  X  30  X  6  =  346  pounds. 

Lathe  Centers. 

In  this  country  lathe  centers  are  universally  made  60  de- 
grees, but  in  Europe  the  most  common  practice  is  to  make  lathe 
centers  90  degrees. 

Morse  Taper. 

The  Morse  Taper,  which  is  so  universally  used  for  the 
shanks  of  drills  and  other  tools,  is  given  in 

TABLE  No.  67.— Horse  Taper. 


No.  of 
Taper. 

Standard 
Plug  Depth. 

Diameter  of 
Plug  at 
Large  End. 

Diameter  of 
Plug  at 
Small  End. 

Taper  per 
Foot. 

1 

2ys  inch. 

0.475   inch. 

0.369   inch. 

0.600  inch. 

2 

9   9          <* 

0.7          " 

0.572      « 

0.602      " 

3 

Q    3          it 
^16 

0.938      « 

0.778      " 

0.602      " 

4 

1.231      " 

1.02 

0.623      « 

5 

5T\     " 

1.748      " 

1.475      " 

0.630      " 

6 

iy4    " 

2.494      « 

2.116      " 

0.626      « 

For  very  complete  information  regarding  the  Morse  Taper, 
see  American  Machinist,  May  14,  1896. 

Jarno  Taper. 

Inthe"Jarno  Taper"  the  number  of  the  taper  gives  the 
length  of  the  standard  plug  in  half-inches,  and  it  gives  the  diame- 
ter of  the  small  end  in  tenths  of  inches  and  the  diameter  of  the 
large  end  in  eighths  of  inches.  For  instance,  a  No.  8  "Jarno 
Taper  "  is  four  inches  long,  one  inch  diameter  at  large  end,  and 
0.8  inch  diameter  at  small  end.  The  taper,  of  course,  is  0.6 
inch  per  foot  for  all  numbers.  This  is  a  very  convenient  sys- 
tem, and  deserves  adoption  for  its  merits.  The  same  taper  is 


SHOP   NOTES. 


429 


also  very  well  adapted  to  the  metric  system,  as  0.6  inch  per 
foot  is  equal  to  0.05  millimeter  per  millimeter. 

The  following  table  is  given  to  illustrate  the  system.  The 
table  could  be  extended  to  as  large  size  tapers  as  are  required 
for  any  work. 

TABLE  No.  68.— Jarno  Taper. 


Number  of 
Taper. 

Length  of 
Taper. 

Diameter  of 
Large  End  of 
Taper. 

Diameter  of 
Small  End  of 
Taper. 

1 

Y* 

^  =  0.125 

A  =  0.1 

2 

1 

%  =  0.250 

J   =0.2 

3 

l/^ 

3^  —  0.375 

A  =  0.3 

4 

2 

y2  =  0.500 

A  =  0.4 

5 

2^ 

5/6  =  0.625 

#  =0.5 

6 

3 

Y±  =  0.750 

1   =0.6 

7 

3^ 

7^  =  0.875 

A  =  o-7 

8 

4 

1  =  1.000 

*   =0.8 

9 

4>^ 

1#  =  1.125 

T9*  =  0.9 

10 

5 

li^  =  1.250 

1      =  1.0 

This  system  of  taper  is  described  by  "Jarno  "  in  the  Amer- 
ican Machinist,  October  31, 1889. 

Marking  Solution. 

Dissolve  one  ounce  of  sulphate  of  copper  (blue  vitriol)  in 
four  ounces  of  water  and  half  a  teaspoonful  of  nitric  acid. 
When  this  solution  is  applied  on  bright  steel  or  iron,  the  surface 
immediately  turns  copper  color,  and  marks  made  by  a  sharp 
scratch-awl  will  be  seen  very  distinctly. 

A  Cheap  Lubricant  for  Milling  and  Drilling. 

Dissolve  separately  in  water  10  pounds  of  whale-oil  soap  and 
15  pounds  of  sal-soda.  Mix  this  in  40  gallons  of  clean  water. 
Add  two  gallons  of  best  lard  oil,  stir  thoroughly,  and  the  solu- 
tion is  ready  for  use. 

Soda  Water  for  Drilling. 

Dissolve  three-fourths  to  one  pound  of  sal-soda  in  one  pail 
full  of  water. 

Solder. 

Ordinary  solder  is  an  alloy  consisting  of  two  parts  of  tin 
and  one  part  of  lead,  and  melts  at  360°. 

Solder  consisting  of  two  parts  of  lead  and  one  part  of  tin 
melts  at  475°.  For  tin  work  use  resin  for  a  flux. 


43°  SHOP   NOTES. 

Soldering  Fluids. 

Add  pieces  of  zinc  to  muriatic  acid  until  the  bubbles  cease 
to  rise,  and  the  acid  may  be  used  for  soldering  with  soft  solder. 

Mix  one  pint  of  grain  alcohol  with  two  tablespoonfuls  of 
chloride  of  zinc.  Shake  well.  This  solution  does  not  rust  the 
joint  as  acids  are  liable  to  do. 

When  soldering  lead  use  tallow  or  resin  for  a  flux,  and  use  a 
solder  consisting  of  one  part  of  tin  and  \yz  parts  of  lead. 

Spelter. 

Hard  spelter  consists  of  one  part  of  copper  and  one  part  of 
zinc. 

A  softer  spelter  is  made  from  two  parts  of  copper  and  three 
parts  of  zinc. 

A  spelter  which  will  flow  very  easily  at  low  heat  consists  of 
46%  of  Copper,  46%  of  Zinc,  and  8%  of  Silver.  When  making 
any  of  these  different  kinds  of  spelter,  melt  the  copper  first  in  a 
black  lead  crucible  and  then  put  in  the  zinc  after  the  copper 
has  cooled  enough  to  furnish  just  sufficient  heat  to  melt  the 
zinc,  but  not  enough  to  burn  it.  Stir  with  an  iron  rod  and 
after  the  metals  have  compounded  and  the  compound  is  still 
molten,  pour  upon  a  basin  of  water.  The  metal  in  striking  the 
water  will  form  into  small  globules  or  shot  and  will  so  cool, 
leaving  a  coarse  granular  spelter  ready  for  use.  When  pouring 
the  metal  let  a  helper  keep  stirring  the  water  with  an  old  broom. 

Alloy  Which  Expands  in  Cooling. 

Melt  together  nine  pounds  of  lead,  two  pounds  of  antimony 
and  one  pound  of  bismuth.  This  alloy  may  be  used  in  fastening 
foundation  bolts  for  machinery  into  foundation  stones.  In  such 
cases,  collars  or  heads  are  left  on  the  bolts  and  after  the  hole 
is  drilled  in  the  stone  a  couple  of  short,  small  holes  are  drilled 
at  an  angle  to  the  big  hole ;  when  the  metal  is  poured  in,  it  will 
flow  around  the  bolts  and  also  into  these  small  holes,  and  it  is 
almost  impossible  for  the  bolt  to  pull  out. 

CAUTION. — When  drilling  holes  in  stone,  water  is  always 
used,  but  this  must  be  carefully  dried  out  by  the  use  of  red-hot 
iron  rods  before  the  melted  metal  is  poured  in.  If  this  pre- 
caution is  not  taken  the  metal  will  blow  out,  making  a  poor  job, 
and  it  may  also  cause  accident  by  burning  the  hands  and  face 
of  the  man  who  is  pouring  it  in. 

Shrinkage  of  Castings. 

General  rule : 

%  inch  per  foot  for  iron. 

S/IQ  inch  per  foot  for  brass. 

In  small  castings  the  molder  generally  raps  the  pattern 
more  than  the  casting  will  shrink,  therefore  no  shrinkage  is  al- 
lowed. Frequently  castings  are  of  such  shape  that  the  pressure 


SHOP   NOTES.  431 

of  the  fluid  iron  on  some  part  of  the  mould  is  liable  to  make  the 
sand  yield  a  little  and  thereby  cause  the  casting  to  be  as  large 
as,  or  even  larger  than  the  pattern.  All  such  things  a  practical 
pattern  maker  takes  into  consideration  when  allowing  for 
shrinkage  in  patterns. 

Case  Hardening  Wrought  Iron  and  Soft  Steel. 

Bone  dust  specially  prepared  for  the  purpose,  or  burnt 
leather  scrap,  is  placed  in  a  cast-iron  box,  together  with  the 
article  to  be  hardened.  Cover  the  top  of  the  box  with  plenty  of 
the  hardening  material  in  order  to  keep  the  air  out.  Heat  the 
whole  mass  slowly  in  a  furnace  to  a  red  heat  from  two  to  five 
hours  in  order  that  it  may  be  uniformly  and  thoroughly  heated 
through.  A  few  iron  rods  about  %<?  inch  in  diameter  are  put  in 
when  packing  the  box,  one  end  of  the  rod  reaching  about  to  the 
middle  of  the  box,  and  the  other  end  projecting  out  through  the 
hardening  material  on  top.  When  the  box  appears  to  have  the 
right  heat,  these  rods  are  pulled  out  one  at  a  time,  in  order  to 
judge  of  the  heat  in  the  center  of  the  mass.  When  the  box 
has  been  exposed  to  the  fire  the  desired  length  of  time,  its 
contents  are  quickly  dumped  into  cool  water. 

Sieves  of  iron  netting  are  laid  on  the  bottom  of  the  tub  into 
which  the  case  hardening  material  is  dumped  so  that  the  hard- 
ened articles  may  be  conveniently  taken  up  from  the  water  by 
one  of  the  sieves.  The  case  hardening  material  itself  is  also 
taken  out  by  another  sieve  which  is  of  very  fine  netting  and 
placed  under  the  first  one.  The  material  is  dried  and  used  over 
again,  and  a  little  new  material  is  added  when  repacking  the 
boxes. 

When  articles  are  well  finished  before  hardening,  this  pro- 
cess gives  a  very  fine  color  to  both  soft  steel  and  wrought  iron. 

Case  hardening  may  also  be  effected  by  packing  the  articles 
in  soot,  but  this  process  does  not  give  a  nice  color. 

Horn  and  hoof  is  also  used  for  case  hardening.  Malleable 
iron  may  also  be  case  hardened,  but  it  requires  careful  handling 
in  order  to  prevent  its  cracking  and  twisting  out  of  shape. 

Case  Hardening  Boxes 

are  made  from  cast-iron  and  are  of  various  sizes.  Small  boxes 
may  be  made  nine  inches  long,  five  inches  wide,  and  four  inches 
deep,  and  about  one-fourth  inch  thick.  They  should  be  pro- 
vided with  legs  at  least  one  inch  high  so  that  the  heat  may  get 
under  the  bottom  as  at  the  top.  An  ear  having  a  rectangular 
hole  through  it  should  be  cast  under  the  bottom  at  each  end  of 
the  box.  This  gives  a  chance  to  handle  the  box  with  a  fork 
having  flat  prongs  instead  of  taking  it  out  of  the  hardening  fur- 
nace with  a  pair  of  tongs,  which  is  liable  to  break  the  box, 
as  cast-iron  is  very  inferior  in  strength  when  hot. 


432  SHOP  NOTES. 

To  Harden  with    Cyanide  of  Potassium. 

Heat  the  cyanide  of  potassium  in  a  wrought  iron  pot  until 
cherry  red,  and  keep  it  so  by  a  steady  fire,  immerse  the  pieces 
to  be  hardened  from  three  to  five  minutes,  according  to  their 
size  and  degree  of  hardness  required,  then  plunge  into  cold 
water.  Large  pieces  require  more  time  than  small  ones,  and 
the  longer  the  article  remains  in  the  cyanide  the  deeper  the 
hardening  becomes.  New  cyanide  gives  the  best  color  and 
cyanide  previously  used  for  hardening  produces  a  harder  surface. 


BLUE    PRINTING. 
To  Prepare  Blue  Print  Paper. 

Dissolve  two  ounces  of  citrate  of  iron  and  ammonium  in  8# 
ounces  of  soft  water.  Keep  this  in  a  dark  bottle.  Also  dissolve 
1%  ounces  of  red  prussiate  of  potash  in  8X  ounces  of  water  and 
keep  in  another  dark  bottle.  When  about  to  use,  mix  (in  a 
dark  place)  an  equal  quantity  of  each  solution  in  a  cup  and 
apply  with  a  sponge  or  a  camel's  hair  brush  as  evenly  as  pos- 
sible' on  one  side  of  white  rag  paper  (such  as  used  for  envelopes). 
Let  it  dry  and  put  it  away  in  a  dark  drawer.  The  paper  must 
not  be  prepared  in  daylight  but  when  taking  prints  it  may  be 
handled  then,  providing  care  is  used  to  expose  it  as  little  as 
possible  to  the  light  before  it  is  put  into  the  printing  frame. 

Blue  Print  Frame. 

Make  a  strong  frame  similar  to  a  picture  frame  having  a 
strong  and  thick  glass.  Make  a  loose  back,  from  boards  about 
y2  inch  thick,  which  is  held  into  the  frame  by  four  suitable 
catches  so  arranged  that  they  press  this  back  firmly  and  evenly 
against  the  glass.  The  surface  next  to  the  glass  should  be 
covered  by  three  thicknesses  of  flannel  in  order  to  make  a 
cushion  so  that  the  prepared  paper  and  the  tracing  are  kept  close 
together  when  put  in  the  frame. 

Blue  Printing. 

The  drawing  must  be  made  on  transparent  material,  for 
instance,  tracing  cloth  or  tracing  paper.  Place  the  tracing  in 
the  frame  with  the  side  on  which  the  drawing  is  made  next  to 
the  glass.  Place  the  prepared  side  of  the  sensitive  paper  against 
the  back  of  the  tracing.  Put  the  loose  back  into  the  frame  with 
the  padded  side  against  the  prepared  paper,  and  fasten  it  up  so 
that  both  paper  and  tracing  are  kept  firmly  against  the  glass. 
Expose  to  sunlight  from  three  to  six  minutes,  according  to  the 
brightness  of  the  sun.  Take  the  sensitive  paper  out  of  the 
frame  and  quickly  put  it  into  a  tub  of  clean  cool  water  and  wash 
it  off,  and  the  drawing  will  appear  in  white  lines  on  blue 
ground.  Hang  the  print  up  by  one  edge  so  that  the  water  will 
run  off,  and  let  the  print  hang  until  dry. 


INDEX. 


Acceleration  due  to  gravity,  276. 
Addition,  5. 

of  decimals,  14. 

of  vulgar  fractions,  12. 

of  logarithms,  75. 
Algebra,  notes  on,  63. 
Allowable  deflection,  266. 
Alloy  which  expands  in  cooling, 

43°- 

Animal  powrer,  318. 
Angular  velocity,  299. 
Area  of  circles,  196. 

tables  of,  209. 
Area  of  segments  of  circles,  199. 

of  parallelograms,  196. 

of  triangles,  193. 

of  triangles,  formulas  for, 
176-177. 

of  trapezoids,  196. 

of  a  circular  lune,  202. 

of  a  zone,  202. 

B. 

Beams,  deflection  in,  254-266. 

transverse  strength  of,  233- 
254. 

to  calculate  size  to  carry  a 
given  load,  247. 

round  wooden,  251. 

not  loaded  at  the  center  of 
span,  252. 

loaded  at  several  places,  253. 

placed  in  an  inclined  posi- 
tion, 254. 


Bearings,  369-374. 

area  of,  369. 

allowable  pressure  in,  369. 

proportions  of,  370-374. 
Belts,  326-338. 

arc  of  contact  of,  332. 

angle,  337. 

cementing,  328. 

lacing,  327. 

length  of,  329. 

horse-power  transmitted  by, 

329- 

crossed,  336. 
oiling  of,  338. 
quarter  turn,  336. 
slipping  of,  337. 
tighteners  on,  337. 
velocity  of,  337. 
Bevel  gears,  389. 

dimensions    of    tooth   parts 

in,  391. 

Body  projected  at  an  angle  up- 
ward, 279. 
in  horizontal  direction  from 

an  elevated  place,  284. 
Blue  print  paper,  how  to  prepare, 

432. 
Blue  printing,  432. 

C. 

Case  hardening,  431. 
boxes,  431. 
by    cyanide    of    potassium, 

432. 

Cap  screws,  407. 
Center  of  gravity,  292. 


433 


434 


INDEX. 


Center  of  gyration,  292. 

of  oscillation,  292. 

of  percussion,  292. 
Centrifugal  force,  302. 
Chain  Links,  323. 
Circles,  152. 

area  and  circumference  of, 

209. 

Couplings,  369. 
Coupling  bolts,  406. 
Compound  proportions,  17. 
Cone,  surface  of,  204. 

volume  of,  205. 
Constant  for  deflection,  how  to 

find,  264. 

Crane  Hooks,  323. 
Cylinder,  thickness  of,  219. 

surface  of,  203. 

volume  of,  204. 
Cube  root,  30. 

table  of,  33. 
Crushing  strength,  223. 
Clearance,  427. 

D. 

Debt  paying  by  instalments,  89. 
Deflection  in  beams  when  loaded 

transversely,  254. 
allowable,  266. 
Derrick,  proportions  of  a  two  ton, 

324. 

Difference  between  one   square 
foot  and  one  foot  square, 

193- 
Dimensions   of   U.   S.  standard 


screws,  405. 
of       Whitword 


standard 


screws,  404. 
Diameter  of  tap-drill,  404. 

of     screws    at    bottom     of 

thread,  403. 
Discount  or  rebate,  83. 
Division,  6. 

of  decimal  fractions,  15. 
of  vulgar  fractions,  10. 
of  logarithms,  78. 
Drawing,  problems  in   geometri- 
cal, 184. 


E. 

Efficiency  of  machinery,  322. 
Ellipse,  area  of,  208. 

circumference  of,  208. 
Elongation  under  tension,  215. 
Energy,  kinetic,  287. 
Eye  bolts,  407. 
Equations,  65. 

quadratic,  67. 
Equation  of  payments,  27. 

P. 

Factor  of  Safety.,  274. 

Force  energy  and  power,  285. 

of  a  blow  how  to  calculate, 
288. 

acceleration      and     motion 
formulas  for,  288. 

centrifugal,  302. 
Formulas,  3. 
Fractions,  addition  of,  7,  14. 

subtraction  of,  8,  14. 

multiplication  of,  9,  15. 

division  of,  10,  15. 

to  reduce  from  one  denom- 
ination to  another,  n. 

to   reduce   a   decimal   to   a 
vulgar,  13. 

to  reduce  a  vulgar  to  a  deci- 
mal, 12. 
Friction,  303. 

axle,  305. 

angle  of,  306. 

co-efficient  of,  304,  306. 

rolling,  304. 

rules  for,  306. 

in  machinery,  306. 

in  pulley-blocks,  307. 
Frustum  of  a  cone,  205. 

surface  of,  205. 

volume  of,  206. 
Frustum  of  a  pyramid,  207. 

surface  of,  207. 

volume  of,  207. 
Fly  wheels,  356. 

weight  of,  356. 

centrifugal  force  in,  357. 


INDEX. 


435 


Fly  Wheels,  to  calculate  speed  of 
bursted,  284. 

safe  speed  of,  360. 

safe  diameter  of,  360. 
French  standard  screws,  409. 

G. 

Gage  for  sheet  iron  and   sheet 

steel,  145. 
for  wire,  146. 
for  twist  drills  and  steel  wire, 

148. 

for  stubs,  148. 
Geometry.,  149-152. 
Geometrical  mean,  70. 
Gravity,  276. 

specific,  138. 
Gear  teeth,  375-400. 

circular  pitch  of,  375. 
comparing  circular  and   di- 
ametral pitch  of,  380. 
cycloid  form  of,  382. 
dimensions  of  teeth,  380. 
diametral  pitch  of,  377. 
involute  form  of,  385. 
width  of,  389. 
Gearing  bevel,  389. 

how  to  calculate    speed  of, 

322. 
dimensions   of    tooth    parts 

in,  380. 
dimensions   of    tooth   parts 

in  bevel,  391. 
worm,  395. 
dimensions   of    tooth   parts 

in  worm,  397. 
elliptical,  400. 

Grindstone,  weight  of,  140,  428. 
speed  of,  321. 


Hauling  a  load,  318. 
Horse  power,  317. 

of  a  steam  engine,  317. 

compound  or  triple  engine, 

3*7. 
of  waterfalls,  318. 


Hot  water  heating,  422. 
Hydraulic,  notes  on,  413. 
Hyperbolic  logarithms,  71,  126, 


I. 


I  beam,  strength  of,  242. 
Impulse,  286. 

Interest,    compound,    computed 
annually,  22. 

semi-annually,  24. 

simple,  19. 

tables,  20-25. 

by  logarithms,  81. 
Impulse,  286. 
Involute,  192. 
Inertia,  285. 

moment  of,  293. 
Inclined  plane,  308. 
International    standard    thread, 
410. 

J. 

Jarno  taper,  428. 
table  of,  429. 


Keys,  proportions  of,  368. 
Kinetic  energy,  287. 

L. 

Lag  screws,  408. 
Lathe  centers,  428. 
Laws,  Newton's,  276. 
Logarithms,  71. 

table  of,  90. 

hyperbolic,  71,  126. 

table  of,  126. 
Levers,  292. 
Lune,  circular,  202. 
Lubricant  for   milling  and  drill- 
ing, 429. 

M. 

Machinery,  efficiency  of,  322. 
power  required  to  drive,  319. 


INDEX. 


Machinery,  speed  of,  320. 
Mathematics,  notes  on,  i. 
Manila  ropes,  transmission  of 

power  by,  344. 
transmission     capacity      of, 

345- 

preservation  of,  347. 

weight  of,  346. 

strength  of,  222. 
Marking  solution,  429. 
Mass,  286. 
Multiplication,  6. 

of  decimals,  15. 

of  vulgar  fractions,  9. 

of  logarithms,  77. 
Mechanics,  276. 
Mensuration,  193. 
Metric    system   of   weights  and 
measures,  135. 

thread    with    inch     divided 
lead-screw,    how    to    cut, 
411,  412. 
Modulus  of  elasticity,  213-216. 

how  to  calculate,  213,  255, 

265. 

Moments,  292. 
Moment  of  inertia,  293. 

polar,  297. 
Momentum,  286. 
Morse  taper,  428. 
Motion,  Newton's  laws  of,  276. 

down  an  inclined  plane,  283. 

N. 

Napierian  logarithms,  71,  126. 
Newton's  laws  of  motion,  276. 

O. 

Oiling  of  belts,  338. 

P. 

Parallelogram  of  forces,  316. 
Partnership,  27. 
Payments,  equation  of,  27. 
Pillars,  safe  load  on  226-229. 
hollow  cast  iron,  228. 


Pillars,  wrought  iron,  231. 

weight  of,  228,  232. 
Polar  moment  of  inertia,  297. 
Polygons,  149. 
Posts,  wooden,  231. 
Power,,  animal,  318. 

of  man,  319. 

required  to  drive  machinery, 

3*9- 
Pulley-blocks,  307. 

differential,  308. 
Pulleys,  348. 

how  to  calculate  size  of,  321. 

stepped,  350. 

correct  diameter  of,  352. 

for  back-geared  lathes,  354. 
Pyramids,  frustum  of  a,  207. 

slanted  area  of  a,  206. 

total  area  of  a,  206. 

volume  of  a,  206. 
Pressure  on  bearings  caused  by 

the  belt,  333. 

Pressure  of  fluid  in  a  vessel,  413. 
Produce,  weight  of,  135. 
Problems   in   geometrical   draw- 
ing, 184. 

Progressions,  68. 
Proportion,  16. 

compound,  17. 

Q. 

Quadratic  equations,  67. 
Quantity  of  water  discharged  by 

pipes,  417. 
of  water   required  to  make 

any     quantity   of    steam, 

422. 

R. 

Radius  of  gyration,  293. 
Radical      quantities      expressed 
without   the  radical   sign, 

32- 

Ratio,  1 6. 
Reciprocals,  32. 
table  of,  33. 
Results  of  small  savings,  26. 


INDEX. 


437 


Rope,  manila,  344,  345,  347*  346, 

and  222. 
wire,  338,  340,  342,  343»  and 

222. 

s. 

Safety,  factor  of,  274. 
Savings,  results  of  small,  26. 
Sector  of  a  circle,  area  of,  198. 
Segmemt  of  a  circle,  199. 

length  of  arc  of,  199. 

area  of,  199,  201. 

of  a  sphere,  volume  of,  208. 
Sinking  funds  and  savings,  84. 
Soda-water  for  drilling,  429. 
Solder,  429. 
Soldering  fluid,  430. 
Subtraction,  5. 

of  decimal  fractions,  14. 

of  vulgar  fractions,  8. 

of  logarithms,  76. 
Screws,  311,  401. 

"  pitch,"    "  inch    pitch,"    of 
worms  and,  401. 

French  standard,  409. 

German  standard,  410. 

international  standard,  410, 
411. 

U.  S.  standard,  402. 

Whitworth  standard,  404. 

diameter  at  bottom  of  thread 
of,  403. 

round  and  fillister  head,  406. 

cap,  407. 

Screw-cutting     by     the     engine 
lathe,  401. 

multiple  threaded,  401. 
Shafting,  360. 

allowable  deflection  in,  362. 

classification  of,  365. 

horse  power  of,  365,  366. 

not  loaded  at  the  middle  be- 
tween bearings,  361. 

torsional  strength  of,  363. 

torsional  deflection  of,  364. 

transverse  strength  of,  360. 

transverse  deflection  of,  361. 
Shafts  for  idlers,  367. 


Shearing  strength,  272. 
Shop  notes,  428-432. 
Shrinkage  of  castings,  430. 
Spelter,  430. 
Speed  of  machinery,  320. 

of  bandsaws,  320. 

of    drilling-machines    (ircjn), 
320. 

of  emery-wheels  and  straps, 
321. 

of  grindstones,  321. 

of  lathes  (for  iron  and  wood), 
320. 

of  milling  machines,  320. 

of  planers,  320. 
Specific  gravity,  138. 
Sphere,  surface  of  a,  207. 

volume  of  a,  208. 
Steam,  notes  on,  418-427. 

properties  of  saturated,  419 

expansion  of,  425-426. 

weight   of,  required  to  boil 
water,  424. 

heating,  421. 
Strength  of  materials,  213-275. 

tensile  strength,  213. 

crushing,  223. 

transverse,  233. 

torsional,  266. 

shearing,  272. 

cast  iron,  214. 

wrought  iron,  214. 

cylinders,  219. 

flat  cylinder  heads,  220. 

of    dished    cylinder    heads, 
220. 

of  bolts,  218. 

of  hollow  sphere,  221. 

of  chains,  222. 

of  wire  rope,  222. 

of  manila  rope,  222. 

of  beams  when    section   is 
not  uniform,  243. 

of    square    and    rectangular 

wooden  beams,  245. 
Square  root,  29. 

table  of,  33. 

by  logarithms,  78. 


INDEX. 


T. 

Tank,    to    calculate   number    of 

gallons  in  a,  202. 
Tensile  strength,  216. 
Torsional  deflection,  271. 

in  hollow  round  shafts,  271. 

strength,  266. 

strength    in    hollow    round 

shafts,  270. 
Transverse  strength,  233. 


Upward  motion,  279. 

U.  S.  Standard  screws,  402. 

V. 

Value  of  various  metals,  139. 
of  the  trigonometrical  func- 
tions for  some  of  the  most 
common  angles,  156. 
low  pressure  steam  for  heat- 

ing purposes,  422. 
Velocity,  angular,  299. 
of  efflux,  413-414. 
of  water  in  pipes,  415. 

W. 

Water,  measure  of,  138. 
weight  of,  138. 


Water  required  to  condense  one 

pound  of  steam,  423. 
Weights  and  measures,  133. 

metric  system  of,  135. 
Weight  of  iron,  square  or  round, 

flat,  144. 

of  any  shape  of  section,  141. 

of  sheet-iron  of   any   thick- 
ness, 141. 

of  steel,  142. 

of  casting  from  weight  of  pat- 
tern, how  to  calculate,  141. 

metals    not     given    in    the 
tables,  141. 

of  cast-iron  balls,  142. 

and  value  of  metals,  139. 

of  various  materials,  140. 

of  liquids,  140. 
Wire  gages,  146. 
Wire  rope  transmissions,  338. 

capacity  of,  340. 

deflection  in,  342. 

weight  of,  343. 

strength  of,  343. 


Zone  circular,  202. 


CLASSIFIED   CATALOGUE   OF 

BOOKS  ON  STEAM,  STEAM  ENGINES,  Ete. 

FOR    SAI.K    I'.Y 

13.     VAN      NOSXRAND      COMPANY, 

23  MURRAY  AND   27  WARREN   STS.,  NEW  YORK. 


BOILERS. 

Barr.  Practical  Treatise  on  High  Pressure  Steam  Boilers,  including 
Results  of  Recent  Experimental  Tests  of  Boiler  Material,  etc.  8vo.  Illus- 
trated. Indianapolis,  1893.  $3.00 

Barrus.  Boiler  Tests  :  Embracing  the  results  of  one  hundred  and  thirty- 
seven  evaporative  tests,  made  on  seventy-one  boilers,  conducted  by  the 
author.  8vo.  Boston,  1895.  $5-oo 

Christie.  Chimney  Design  and  Theory.  A  book  for  Engineers  and  Archi- 
tects, containing  all  data  relative  to  Chimney  Designing.  Illustrated 
with  numerous  diagrams  and  half-tone  cuts  of  many  famous  chim- 
neys. 8vo,  cloth.  Illustrated.  New  York,  1899.  $3.00 

Courtney.  The  Boiler  Maker's  Assistant  in  Drawing,  Templating,  and 
Calculating  Boiler  Work  and  Tank  Work,  with  rules  for  the  Evapora- 
tive Power  and  the  Horse  Power  of  Steam  Boilers,  and  the  Proportions 
of  Safety  Valves,  and  Useful  Tables  of  Rivet  Joints  of  Circles,  Weights 
of  Metals,  etc.  Revised  and  edited  by  D.  K.  Clark,  C.E.  Illustrated. 
London,  1898.  (Weale's  Series.)  $0.80 

The  Boiler  Maker's  Ready  Reckoner.  With  examples  of  Practical 
Geometry  and  Templating,  for  the  Use  of  Platers'  Smiths,  and  Riveters. 
Revised  and  edited  by  D.  K.  Clark.  3d  edition.  London,  1890.  (Weale's 
Series.)  #1.60 

Davis.  A  Treatise  on  Steam-Boiler  Incrustation,  and  Methods  for  Pre- 
venting Corrosion  and  the  Formation  of  Scale  ;  also  a  Complete  List 
of  all  American  Patents  issued  by  the  Government  of  the  United  States 
from  1790  to  July  I,  1884,  for  Compounds  and  Mechanical  Devices  for 
Purifying  Water,  and  for  Preventing  the  Incrustation  of  Steam  Boilers. 
65  engravings.  8vo.  Philadelphia,  1884.  $2.00 

Foley,  Nelson.  The  Mechanical  Engineer's  Reference  Book  for  Ma- 
chine and  Boiler  Construction,  in  two  parts.  Part  I.,  General  En- 
gineering Data.  Part  II.,  Boiler  Construction.  With  51  Plates  and 
numerous  illustrations  specially  drawn  for  this  work.  Folio,  half  mor. 
London,  1895.  $25.00 

Horner.  Plating  and  Boiler  Making.  A  Practical  Handbook  for  Work- 
shop Operation,  including  an  Appendix  of  tables  by  A  Foreman  Pattern 
Maker.  338  illustrations.  12 mo.  London,  1895.  $3.00 


LIST  Of  BOOKS. 

Button.  Steam  Boiler  Construction:  A  Practical  Handbook  for  Engi- 
neers, Boiler  Makers,  and  Steam  Users.  With  upwards  of  300  illustra- 
tions. 3d  edition.  8vo.  London,  1898.  $6.00 

Munro.  Steam  Boilers :  Their  Defects,  Management,  and  Construction. 
2d  edition  enlarged,  with  numerous  illustrations  and  tables.  I2mo. 
London,  1892.  $1.50 

Roper.  The  Steam  Boiler:  Its  Care  and  Management.  With  instruc- 
tions for  increasing  the  Efficiency  and  Economy,  and  insuring  the  Dura- 
bility and  Longevity  of  all  classes  of  Steam  Boilers,  Stationary,  Loco- 
motive, Marine,  and  Portable.  With  Hints  and  Suggestions  and  Advice 
to  Engineers,  Firemen,  and  Owners  of  Steam  Boilers.  4th  edition, 
revised.  I2mo,  tuck,  mor.  Philadelphia,  1897.  $2.00 

Use  and  Abuse  of  the  Steam  Boiler.  Illustrated,  nth  edition. 

I2mo,  mor.  tucks.  Philadelphia,  1897.  $2.00 

Rose.  Steam  Boilers.  A  Practical  Treatise  on  Boiler  Construction  and 
Examination.  For  the  Use  of  Practical  Boiler  Makers,  Boiler  Users, 
and  Inspectors,  and  embracing  in  plain  figures  all  the  calculations  neces- 
sary in  designing  and  classifying  Steam  Boilers.  73  engravings.  8vo. 
Philadelphia,  1897.  $2-5o 

Rowan.  On  Boiler  Incrustation  and  Corrosion.  New  edition,  revised 
and  enlarged  by  F.  E.  Idell.  i6mo,  boards.  New  York,  1895.  $0.50 

Sexton.  Pocket  Book  for  Boiler  Makers  and  Steam  Users,  comprising  a 
variety  of  useful  information  for  Employer  and  Workman,  Government 
Inspectors,  Board  of  Trade  Surveyors,  Engineers  in  charge  of  Works 
and  Slips,  Foremen  of  Manufactories,  and  the  General  Steam-Using 
Public.  4th  edition,  revised  and  enlarged.  32010,  roan.  London,  1895. 

$2.00 

Stromeyer.  Marine  Boiler  Management  and  Construction.  Being  a 
Treatise  on  Boiler  Troubles  and  Repairs,  Corrosions,  Fuels,  and  Heat. 
On  the  Properties  of  Iron  and  Steel,  on  Boiler  Mechanics,  Workshop 
Practices,  and  Boiler  Designs.  8vo.  London,  1893.  $5.00 

Thurston.  Manual  of  Steam  Boilers  :  Their  Designs,  Construction,  and 
Operation.  For  Technical  Schools  and  Engineers.  183  engravings  in 
text.  6th  edition,  8vo.  New  York,  1898.  $5-oo 

Steam  Boiler  Explosions.  In  Theory  and  Practice,  Illustrated. 

2d  edition,  I2mo.  New  York,  1888.  $1.50 

A  Handbook  of  Engine  and  Boiler  Trials,  and  of  the  Indicator  and 

Prony  Brake,  for  Engineers  and  Technical  Schools.  3d  edition.  Illus- 
trated, 8vo.  New  York,  1897.  #5.00 


D.    VAN  NOSTRAND   COMPANY. 

Traill.  Boilers:  Their  Construction  and  Strength.  A  Handbook  of 
Rules,  Formulae,  Tables,  etc.,  relative  to  Material,  Scantlings,  and  Pres- 
sures, Safety  Valves,  Springs,  Fittings  and  Mountings,  etc.  For  use  of 
Engineers,  Surveyors,  Draughtsmen,  Boiler  Makers,  and  Steam  Users. 
With  illustrations.  3d  edition,  I2mo,  mor.  London,  1896.  $5.00 

Triplex.  Marine  Boilers.  A  Treatise  on  the  Causes  and  Prevention  of 
their  Priming,  with  Remarks  on  their  General  Management.  Illustrated. 
I2mo.  Sunderland,  1899.  $2.00 

Watson.  Small  Engines  and  Boilers.  A  Manual  of  Concise  and 
Specific  Directions  for  the  Construction  of  Small  Steam  Engines  and 
Boilers  of  Modern  Types  from  five  horse-power  down  to  model  sizes. 
I2mo,  cloth.  Illustrated  with  numerous  diagrams  and  half-tone  cuts. 
New  York,  1899.  $1.25 

The  intention  of  the  author  in  writing  this  work  has  been  to  fur- 
nish specific  directions  and  correct  dimensioned  plans  for  small 
engines  and  boilers. 

Wilson.  A  Treatise  on  Steam  Boilers:  Their  Strength,  Construction, 
and  Economical  Working.  Enlarged  and  illustrated  from  the  Fifth  Eng- 
lish edition  by  J.  T.  Flather.  I2mo.  New  York,  1897.  $2.50 
—  Boiler  and  Factory  Chimneys :  Their  Draught  Power  and  Stability. 
3d  edition,  I2mo.  London,  1892.  #1.50 

FUELS. 

Abbott.  Treatise  on  Fuel.  Founded  on  the  original  Treatise  of  Sir  W. 
Siemens.  Illustrated.  i6mo.  New  York,  1891.  $0.50 

Barr.  Practical  Treatise  on  the  Combustion  of  Coal,  including  descrip- 
tions of  various  mechanical  devices  for  the  Economic  Generation  of 
Heat  by  the  Combustion  of  Fuel,  whether  Solid,  Liquid,  or  Gaseous. 
8vo.  1879.  #2-5° 

Clark  and  Williams.  Fuel :  Its  Combustion  and  Economy,  consisting  of 
Abridgments  of  Treatise  on  the  Combustion  of  Coal  and  the  Economy 
of  Fuel.  With  extensive  additions  in  recent  practice  in  the  Combustion 
and  Economy  of  Fuel,  Coal,  Coke,  Wood,  Peat,  Petroleum,  etc.  4th 
edition.  I2mo.  London,  1891.  $1.50 

Hodgetts.  Liquid  Fuel  for  Mechanical  and  Industrial  Purposes.  Illus- 
trated. Svo.  London,  1890.  $2.50 

Phillips.  Fuels  :  Solid,  Liquid,  and  Gaseous  ;  their  Analysis  and  Valua- 
tion. For  the  use  of  Chemists  and  Engineers,  I2mo.  London,  1896. 

£0.80 


LIST  OF  BOOKS. 

Sexton,  A.  H.  Fuels  and  Refractory  Materials.  Svo.  Cloth.  London, 
1897-  $2.00 

Williams.  Fuel  :  Its  Combustion  and  Economy.  Consisting  of  an 
Abridgment  of  "  A  Treatise  on  the  Combustion  of  Coal  and  the  Pre- 
vention of  Smoke."  With  extensive  additions  by  D.  Kinnear  Clark. 
4th  edition.  London,  1891.  $1.50 

GAS   ENGINES. 

Clerk.  The  Theory  of  the  Gas  Engine.  2d  edition,  with  Additional 
Matter  edited  by  F.  E.  Idell.  i6mo.  New  York,  1891.  $0.50 

The  Gas  Engine.  History  and  Practical  Working.  With  100  illus- 
trations. 6th  edition.  I2mo.  New  York,  1896.  $4.00 

Donkin.  A  Text-Book  on  Gas,  Oil,  and  Air  Engines :  or  Internal  Com- 
bustion Motors  without  Boiler.  154  illustrations.  Svo.  London,  1896. 

$7-5° 

Goodeve.     On  Gas  Engines :  with  Appendix  describing  a  Recent  Engine 

with  Tube  Igniter.      I2mo.     London,  1887.  $1.00 

Hiscox,  Gardner  D.     Gas,  Gasoline  and  Oil  Vapor  Engines.     A  New 

Book  Descriptive  of  Their  Theory  and  Power.    Second  edition,  revised 

and  enlarged.     Svo,  cloth.     Illustrated.     New  York,  1898.  $2.50 

HEAT.  —  THERMODYNAMICS. 

Anderson.  On  the  Conversion  of  Heat  into  Work.  A  Practical  Hand- 
book on  Heat  Engines,  jd  edition.  Illustrated.  I2mo.  London, 
1893-  #2.25 

Box.  Treatise  on  Heat  as  Applied  to  the  Useful  Arts,  for  the  use  of 
Engineers,  Architects,  etc.  8th  edition.  I2mo.  London,  1895.  $5.00 

Larden.     A  School  Course  on  Heat.    Illus.    i2mo.    London,  1894.     $2.00 

McCulloch.  Elementary  Treatise  on  the  Mechanical  Theory  of  Heat  and 
its  application  to  Air  and  Steam  Engine.  Svo.  New  York,  1876.  $3.50 

Maxwell.  Theory  of  Heat.  New  edition,  with  Corrections  and  Addi- 
tions by  Lord  Rayleigh,  Sec.  R.  S.  Illustrated.  I2mo.  New  York, 
1897.  $1.50 

Peabody.  Thermodynamics  of  the  Steam  Engine  and  other  Heat  En- 
gines. Svo.  New  York,  1898.  $5.00 

Rontgen.  The  Principles  of  Thermodynamics.  With  special  Applica- 
tions to  Hot  Air,  Gas,  and  Steam  Engines.  With  additions  from  Profes- 
sors Verdet,  Zeuner,  and  Pernolet.  Translated,  newly  and  thoroughly 
revised  and  enlarged  by  Professor  A.  Jay  Du  Bois.  732  pages,  jd  edi- 
tion. Svo.  New  York,  1896.  $5-oo 


D.    VAN  NOSTRAXD    COMPANY. 

Tyndall.     Heat  considered    as  a    mode  of  Motion.     6th  edition.     I2mo. 

New  York,  1890.  $2.50 

Williams.  On  Heat  and  Steam  :  embracing  New  Views  of  Evaporization, 

Condensation,  and  Expansion.  Illus.  8vo.  Philadelphia,  1882.  $2.50 
Wood.  Thermodynamics,  Heat  Motors,  and  Refrigerating  Machines. 

Revised  and  enlarged  edition.     8vo.     New  York,  1895.  $4.00 

HOISTING    MACHINERY. 

Colyer.  Hydraulic,  Steam  and  Hand  Power-Lifting  and  Pressing  Ma- 
chinery. 72  large  plates.  Svo.  London,  1892.  $10.00 

Glynn.  Treatise  on  the  Construction  of  Cranes  and  other  Hoisting  Ma- 
chinery, yth  edition.  Illustrated.  London,  1887.  $0.60 

Marks.  Notes  on  the  Construction  of  Cranes  and  Lifting  Machinery. 
I2mo.  London.  1892.  $1.00 

Towne.  A  Treatise  on  Cranes,  descriptive  particularly  of  those  designed 
and  built  by  the  Yale  and  Towne  Manufacturing  Company,  owning  and 
operating  the  Western  Crane  Company  ,  including  also  a  description  of 
light  hoisting  machinery  as  built  by  the  same  makers.  Svo.  New  York, 
1883.  $1.00 

Weisbach  and  Hermann.  The  Mechanics  of  Hoisting  Machinery,  in- 
cluding Accumulators,  Excavators,  and  Pile-drivers.  A  Text-book  for 
Technical  Schools  and  a  guide  for  Practical  Engineers.  Authorized  trans- 
lation from  the  second  German  edition  by  Karl  P.  Dahlstrom.  177  illus- 
trations. Svo.  New  York,  1893.  $3-75 

ICE-MAKING    MACHINES. 

Dixon.  Manual  of  Ice-Making  and  Refrigerating  Machines.  A  Treatise 
on  the  Theory  and  Practice  of  Cold-Production  by  Mechanical  Means. 
i6mo.  St  Louis,  1894.  $1.00 

Leask.  Refrigerating  Machinery.  Its  Principles  and  Management. 
With  numerous  illustrations.  Svo.  London,  1894.  $2.00 

Ledoux.  Ice-Making  Machines  :  the  Theory  of  the  Action  of  the  Various 
Forms  of  Cold-producing  or  so-called  Ice-Machines.  Translated  from 
the  French.  248  pages  and  numerous  tables.  i6mo.  New  York,  1892. 

$0.50 

Redwood.  Theoretical  and  Practical  Ammonia  Refrigeration.  A  Prac- 
tical handbook  for  the  use  of  those  in  charge  of  refrigerating  plants. 
Illustrated  with  numerous  Tables.  I2mo.  New  York,  1896.  $1.00 

Wallis-Tayler.  Refrigerating  and  Ice-Making  Machinery.  I2mo,  cloth. 
Illustrated.  London,  1896.  $3-oo 


LIST  OF  BOOKS. 


INDICATORS. 

Bacon.  Treatise  on  the  Richards  Steam  Engine  Indicator.  With  a 
Supplement,  describing  the  latest  Improvements  in  the  Instruments  for 
Taking,  Measuring,  and  Computing  Diagrams.  Also  an  Appendix,  con- 
taining Useful  Formulas  and  Rules  for  Engineers.  23  diagrams.  4th 
edition.  i6mo,  flex.  New  York,  1883.  $1.00 

Ellison.  Practical  Applications  of  the  Indicator.  With  reference  to  the 
Adjustment  of  Valve  Gear  on  all  Styles  of  Engines.  2d  edition.  8vo. 
100  engravings.  Chicago,  1897.  $2.00 

Hemenway.  Indicator  Practice  and  Steam  Engine  Economy.  With 
Plain  Directions  for  Attaching  the  Indicator,  Taking  Diagrams,  Comput- 
ing the  Horse-Power,  Drawing  the  Theoretical  Curve,  Calculating  Steam 
Consumption,  Determining  Economy,  Locating  Derangement  of  Valves, 
and  making  all  desired  deductions ;  also,  Tables  required  in  making  the 
necessary  computations,  and  an  Outline  of  Current  Practice  in  Testing 
Steam  Engines  and  Boilers.  6th  edition.  I2mo.  New  York,  1898. 

$2.00 

Le  Van.  The  Steam  Engine  Indicator  and  its  Use.  A  Guide  to  Practi- 
cal Working  Engineers  for  greater  economy,  and  the  better  Working  of 
Steam  Engines.  i8mo,  boards.  New  York,  1896.  $0.50 

The  Steam  Engine  and  the  Indicator  :  Their  Origin  and  Progressive 

Development,  including  the  most  recent  examples  of  Steam  and  Gas 
Motors,  together  with  the  Indicator,  its  Principles,  its  Utility,  and  its  Ap- 
plication. Illustrated  by  205  engravings,  chiefly  of  Indicator-cards.  8vo. 
Philadelphia,  1890.  $4.00 

Porter.  A  Treatise  on  the  Richards  Steam  Engine  Indicator,  and  the 
Development  and  Application  of  Force  in  the  Steam  Engine.  5th  edi- 
tion, revised  and  enlarged.  8vo.  London,  1894.  $3.00 

Pray.  Twenty  Years  with  the  Indicator.  Being  a  Practical  Text-book 
for  the  Engineer  or  the  Student,  with  no  Complex  Formulae.  With 
many  illustrations  and  rules  as  to  the  best  way  to  run  any  Steam  Engine 
to  get  the  most  economical  results.  How  to  Adjust  Valves  and  Valve 
Motions  Correctly.  Full  directions  for  working  out  Horse-Power,  the 
Amount  of  Steam  or  Water  per  Horse-Power,  Economy  and  Fuel.  Ex- 
tended directions  for  Attaching  the  Indicator,  what  Motions  to  use  and 
those  not  to  use.  Full  directions  for  Computation  of  Power  by  Planim- 
eter  and  other  methods,  with  many  tables  and  hints.  8vo.  New  York, 
1896.  #2.50 


/., 


